1What does C.G. stand for in the study of a plane lamina?
Experimental study of CG location of symmetrical shape
Easy
A.Circle of geometry
B.Curve of gradient
C.Corner of graph
D.Center of gravity
Correct Answer: Center of gravity
Explanation:
C.G. stands for center of gravity, the point through which the weight of an object effectively acts.
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2Where is the C.G. of a uniform rectangular lamina located?
Experimental study of CG location of symmetrical shape
Easy
A.At the midpoint of one side
B.At the midpoint of a diagonal half
C.At one of its four corners
D.At the intersection of its diagonals
Correct Answer: At the intersection of its diagonals
Explanation:
A uniform rectangle has two axes of symmetry, and its C.G. lies at their intersection.
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3Where is the C.G. of a uniform circular lamina located?
Experimental study of CG location of symmetrical shape
Easy
A.At the end of a diameter
B.At the center of the circle
C.At a point on the circumference
D.At the top of the circle
Correct Answer: At the center of the circle
Explanation:
Because a uniform circle is symmetric about every diameter, its C.G. is at its center.
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4Which property helps predict the C.G. of a uniform symmetrical lamina?
Experimental study of CG location of symmetrical shape
Easy
A.Its symmetry axes
B.Its surface color
C.Its hanging time
D.Its material name
Correct Answer: Its symmetry axes
Explanation:
The C.G. of a uniform symmetrical lamina lies on each of its axes of symmetry.
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5Where is the C.G. of a uniform square lamina?
Experimental study of CG location of symmetrical shape
Easy
A.At the upper-left corner
B.At the center of one edge
C.At the lower-right corner
D.At the center of the square
Correct Answer: At the center of the square
Explanation:
The symmetry axes and diagonals of a uniform square meet at its center, which is its C.G.
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6What is commonly attached to the suspension point when locating C.G. experimentally?
Experimental study of CG location of symmetrical shape
Easy
A.A spring balance
B.A plumb line
C.A measuring cylinder
D.A thermometer
Correct Answer: A plumb line
Explanation:
A plumb line shows the vertical direction through the suspension point.
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7When a lamina hangs freely, where does its C.G. lie relative to the suspension point?
Experimental study of CG location of symmetrical shape
Easy
A.Directly below it
B.Directly above it
C.Outside its vertical plane
D.Horizontally beside it
Correct Answer: Directly below it
Explanation:
A freely suspended lamina settles with its C.G. vertically below the suspension point.
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8For a uniform triangular lamina, the C.G. is located at the intersection of its:
Experimental study of CG location of symmetrical shape
Easy
A.Three medians
B.Three side extensions
C.Three perpendicular edges
D.Three exterior angles
Correct Answer: Three medians
Explanation:
The three medians of a uniform triangular lamina meet at its C.G., called the centroid.
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9A uniform lamina has two axes of symmetry. Where should its C.G. be?
Experimental study of CG location of symmetrical shape
Easy
A.At their intersection
B.At its highest point
C.Near its suspension hole
D.On its outer boundary
Correct Answer: At their intersection
Explanation:
The C.G. must lie on both symmetry axes, so it is located where they intersect.
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10What assumption is usually made about a lamina in a simple C.G. experiment?
Experimental study of CG location of symmetrical shape
Easy
A.Its surface is completely frictionless
B.Its edges are always curved
C.Its mass is uniformly distributed
D.Its thickness changes continuously
Correct Answer: Its mass is uniformly distributed
Explanation:
Uniform mass distribution allows geometric symmetry to be used when determining the C.G.
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11Which method is commonly used to find the C.G. of an irregular lamina?
Experimental study of CG location of unsymmetrical shape
Easy
A.Measuring only its perimeter
B.Suspending it from different points
C.Heating it at several points
D.Folding it along one edge
Correct Answer: Suspending it from different points
Explanation:
Suspending the lamina from different points allows vertical lines through its C.G. to be drawn.
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12How many suspension points are normally sufficient to locate the C.G. of an irregular lamina?
Experimental study of CG location of unsymmetrical shape
Easy
A.At least two points
B.Exactly one point
C.At least five points
D.Exactly eight points
Correct Answer: At least two points
Explanation:
Two different suspension points produce two vertical lines whose intersection locates the C.G.
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13What should be done after an irregular lamina stops swinging freely?
Experimental study of CG location of unsymmetrical shape
Easy
A.Measure its horizontal width
B.Remove its suspension hole
C.Rotate it while hanging
D.Trace the vertical plumb line
Correct Answer: Trace the vertical plumb line
Explanation:
The plumb line marks a vertical line that passes through the C.G. of the hanging lamina.
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14What identifies the C.G. after an irregular lamina is suspended from two points?
Experimental study of CG location of unsymmetrical shape
Easy
A.The end of the longer line
B.The intersection of both lines
C.The midpoint of either line
D.The first suspension point
Correct Answer: The intersection of both lines
Explanation:
Each vertical line passes through the C.G., so their intersection gives its position.
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15Why is an irregular lamina suspended from a second point?
Experimental study of CG location of unsymmetrical shape
Easy
A.To change its total mass
B.To make its shape symmetrical
C.To obtain another vertical line
D.To increase its surface area
Correct Answer: To obtain another vertical line
Explanation:
A second vertical line is needed to find the intersection that marks the C.G.
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16Which instrument provides a reliable vertical reference in the suspension experiment?
Experimental study of CG location of unsymmetrical shape
Easy
A.A protractor
B.A meter rule
C.A beam balance
D.A plumb line
Correct Answer: A plumb line
Explanation:
The weight on a plumb line pulls its string vertically downward.
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17Why should the lamina be allowed to stop swinging before a line is drawn?
Experimental study of CG location of unsymmetrical shape
Easy
A.To obtain an accurate vertical line
B.To enlarge the suspension hole
C.To reduce the lamina's area
D.To increase the lamina's weight
Correct Answer: To obtain an accurate vertical line
Explanation:
Waiting until the lamina is stationary makes the plumb-line direction easier to mark accurately.
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18How can the experimentally found C.G. of an irregular lamina be checked?
Experimental study of CG location of unsymmetrical shape
Easy
A.Cut the lamina through that point
B.Measure the color at that point
C.Balance the lamina on that point
D.Heat the lamina at that point
Correct Answer: Balance the lamina on that point
Explanation:
If the marked point is the C.G., the lamina can balance on a narrow support placed there.
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19What may happen if the suspension points are chosen very close together?
Experimental study of CG location of unsymmetrical shape
Easy
A.The C.G. moves outside the lamina
B.The marked lines may nearly overlap
C.The plumb line becomes horizontal
D.The lamina becomes more massive
Correct Answer: The marked lines may nearly overlap
Explanation:
Well-separated suspension points usually produce clearer intersecting lines and improve accuracy.
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20What does the plumb bob do in a C.G. experiment?
Experimental study of CG location of unsymmetrical shape
Easy
A.It indicates the vertical direction
B.It changes the suspension point
C.It stretches the lamina
D.It measures the lamina's area
Correct Answer: It indicates the vertical direction
Explanation:
Gravity pulls the plumb bob downward, making its string a vertical reference line.
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21A uniform rectangular lamina is suspended from one corner, and a vertical line is drawn using a plumb bob. What must be done next to locate its C.G experimentally?
Experimental study of CG location of symmetrical shape
Medium
A.Balance it using the same suspension point again
B.Suspend it from another point and draw a second vertical line
C.Measure the length of the first vertical line
D.Rotate it through on a horizontal table
Correct Answer: Suspend it from another point and draw a second vertical line
Explanation:
The C.G lies on every vertical suspension line, so the intersection of two such lines locates it.
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22A uniform rectangular sheet measures . Taking one corner as the origin, where should its experimentally determined C.G be located?
Experimental study of CG location of symmetrical shape
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
For a uniform rectangle, the C.G is at the intersection of its symmetry axes: cm.
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23A uniform square lamina is suspended from its upper-left corner. Which path will the plumb line follow across the lamina?
Experimental study of CG location of symmetrical shape
Medium
A.A line joining the midpoints of two adjacent sides
B.The vertical edge directly below the suspension corner
C.The diagonal joining the suspension corner to the opposite corner
D.The upper horizontal edge passing through the suspension corner
Correct Answer: The diagonal joining the suspension corner to the opposite corner
Explanation:
The square's C.G is at its center, so the vertical through the suspension corner and C.G follows that diagonal.
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24An experimentally drawn vertical line on a uniform circular disc misses the center marked using its diameter by several millimetres. What is the most likely experimental cause?
Experimental study of CG location of symmetrical shape
Medium
A.The suspension point permanently changes the disc's mass
B.The disc was marked before the plumb bob stopped oscillating
C.The C.G of a circle must lie on its circumference
D.The circular disc has no definite center of gravity
Correct Answer: The disc was marked before the plumb bob stopped oscillating
Explanation:
A uniform disc has its C.G at the geometric center. Marking while the string oscillates can produce an inaccurate vertical.
Incorrect! Try again.
25A uniform isosceles triangular lamina has only its vertical axis of symmetry marked. What can be concluded about its C.G before performing the suspension experiment?
Experimental study of CG location of symmetrical shape
Medium
A.It lies somewhere along the triangle's base
B.It lies at the midpoint of either equal side
C.It lies somewhere on the marked symmetry axis
D.It lies at the midpoint of the triangle's height
Correct Answer: It lies somewhere on the marked symmetry axis
Explanation:
Symmetry fixes the C.G to the symmetry axis, but another construction or suspension is needed to determine its exact position.
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26Two vertical lines obtained by suspending a uniform rectangular lamina intersect away from the intersection of its diagonals. Which conclusion is most reasonable?
Experimental study of CG location of symmetrical shape
Medium
A.A rectangle's C.G always lies near one of its corners
B.The diagonal intersection cannot represent a rectangle's C.G
C.The lamina may be nonuniform or the experiment may contain error
D.Suspension methods apply only to triangular laminas
Correct Answer: The lamina may be nonuniform or the experiment may contain error
Explanation:
For a uniform rectangle, symmetry places the C.G at the diagonal intersection. A mismatch suggests nonuniformity or measurement error.
Incorrect! Try again.
27A uniform equilateral triangular lamina has a height of . How far above its base should the experimentally found C.G lie?
Experimental study of CG location of symmetrical shape
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The centroid of a uniform triangle is one-third of its height above the base, so .
Incorrect! Try again.
28Why is balancing a symmetrical lamina on a narrow knife edge useful after locating its C.G by suspension?
Experimental study of CG location of symmetrical shape
Medium
A.It increases the lamina's number of symmetry axes
B.It moves the C.G onto the lamina's nearest edge
C.It provides an independent check of the located C.G
D.It removes the effect of mass from the experiment
Correct Answer: It provides an independent check of the located C.G
Explanation:
If the support line passes through the C.G, the lamina can balance without a turning effect, confirming the suspension result.
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29A uniform circular lamina is suspended successively from three points on its rim. Ideally, how should the three plumb lines appear?
Experimental study of CG location of symmetrical shape
Medium
A.They should form a smaller circle inside the lamina
B.They should remain parallel without crossing the disc
C.They should all pass through the geometric center
D.They should meet at three separate points on the rim
Correct Answer: They should all pass through the geometric center
Explanation:
Each suspension vertical passes through the C.G, which coincides with the geometric center of a uniform circular lamina.
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30A uniform rectangular lamina balances horizontally when supported at its center. A small equal-sided piece is cut from the left edge only. What will happen to the C.G?
Experimental study of CG location of symmetrical shape
Medium
A.It will shift toward the left side
B.It will remain at the original center
C.It will shift toward the right side
D.It will move vertically outside the lamina
Correct Answer: It will shift toward the right side
Explanation:
Removing mass from the left makes the remaining mass distribution heavier toward the right, shifting the C.G rightward.
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31An irregular cardboard lamina is suspended from point . After it settles, where must its C.G lie?
Experimental study of CG location of unsymmetrical shape
Medium
A.At the lowest point of the cardboard edge
B.On the horizontal line passing through
C.On the vertical line directly below
D.At the midpoint of the cardboard's longest side
Correct Answer: On the vertical line directly below
Explanation:
At equilibrium, the C.G lies vertically below the suspension point so that the weight produces no turning moment.
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32For an irregular lamina, why should two suspension points be chosen well apart rather than very close together?
Experimental study of CG location of unsymmetrical shape
Medium
A.The lamina becomes symmetrical during the experiment
B.Their vertical lines become exactly equal in length
C.Their vertical lines intersect at a clearer angle
D.The plumb bob becomes independent of gravity
Correct Answer: Their vertical lines intersect at a clearer angle
Explanation:
Widely separated suspension points usually produce lines with a larger intersection angle, reducing uncertainty in the C.G position.
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33Two plumb lines drawn on an irregular lamina are represented by and . What are the coordinates of the experimental C.G?
Experimental study of CG location of unsymmetrical shape
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
The C.G is at the lines' intersection. Substituting into gives .
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34Three suspension lines on an irregular lamina form a very small triangle instead of meeting at one point. What is the best estimate of the C.G?
Experimental study of CG location of unsymmetrical shape
Medium
A.The midpoint of the lamina's bounding rectangle
B.The central region of the small triangle
C.The highest vertex of the small triangle
D.The suspension point used for the longest line
Correct Answer: The central region of the small triangle
Explanation:
Small measurement errors can prevent exact concurrence. The central region of the small error triangle is a reasonable estimate.
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35A small lump of clay is attached to the right side of an irregular lamina after its C.G has been marked. Where will the new C.G generally move?
Experimental study of CG location of unsymmetrical shape
Medium
A.Back to the geometric center
B.Toward the attached clay
C.Away from the attached clay
D.Directly toward the suspension hole
Correct Answer: Toward the attached clay
Explanation:
Adding mass on the right increases the weighted mass distribution there, so the combined C.G shifts toward the clay.
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36An irregular crescent-shaped lamina is tested by suspension. The plumb lines intersect in an empty region inside the crescent's curve. Is this result physically possible?
Experimental study of CG location of unsymmetrical shape
Medium
A.No, plumb lines can intersect only at a suspension hole
B.Yes, a C.G can lie outside the material of a concave body
C.Yes, but only when the lamina has uniform symmetry
D.No, a C.G must always lie within the solid material
Correct Answer: Yes, a C.G can lie outside the material of a concave body
Explanation:
The C.G represents the average position of mass and may lie in empty space for a concave or hollow shape.
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37An irregular lamina is rotated before being suspended from the same body-fixed hole. How does its C.G position relative to the lamina change?
Experimental study of CG location of unsymmetrical shape
Medium
A.It remains fixed at the same point on the lamina
B.It shifts toward the new downward-facing edge
C.It moves to the lamina's geometrically lowest point
D.It moves onto the new vertical suspension line permanently
Correct Answer: It remains fixed at the same point on the lamina
Explanation:
The C.G is determined by mass distribution and remains fixed relative to the body, although its position relative to the room changes.
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38A student holds the plumb string slightly sideways while marking a line on an irregular lamina. What effect is most likely?
Experimental study of CG location of unsymmetrical shape
Medium
A.The marked line may not pass through the true C.G
D.The string correctly identifies a horizontal reference
Correct Answer: The marked line may not pass through the true C.G
Explanation:
The string must hang freely under gravity. Sideways contact or force prevents it from showing the true vertical through the C.G.
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39Two small masses form an unsymmetrical model: is at and is at . Where is their combined C.G?
Experimental study of CG location of unsymmetrical shape
Medium
A.
B.
C.
D.
Correct Answer:
Explanation:
Using gives .
Incorrect! Try again.
40An irregular lamina balances on a pencil tip placed at the intersection of its suspension lines. What does this observation confirm?
Experimental study of CG location of unsymmetrical shape
Medium
A.The geometric center and C.G must always be identical
B.The support passes approximately through the lamina's C.G
C.The support has shifted the C.G to the nearest edge
D.The lamina now has equal mass in every small region
Correct Answer: The support passes approximately through the lamina's C.G
Explanation:
Balancing indicates that the resultant weight acts through the support, so the support is approximately beneath the C.G.
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41Three equal-precision suspension trials on a uniform symmetrical lamina produce the traced lines , , and . Using the point that minimizes the sum of squared perpendicular distances to the lines, what is the best estimate of the C.G.?
Experimental study of CG location of symmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Minimizing gives and .
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42A rectangular lamina is geometrically symmetric about both centerlines, but its measured C.G. does not lie on the vertical centerline. Which conclusion is most scientifically justified?
Experimental study of CG location of symmetrical shape
Hard
A.The plumb line method cannot be used for rectangles
B.The C.G. must still be at the diagonal intersection
C.Geometric symmetry alone guarantees a centered C.G.
D.The mass distribution or experiment lacks vertical symmetry
Correct Answer: The mass distribution or experiment lacks vertical symmetry
Explanation:
A geometric symmetry axis contains the C.G. only when the mass distribution shares that symmetry; experimental bias is another possible cause.
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43A uniform isosceles triangular lamina of altitude is suspended in two different orientations. Where should the traced plumb lines intersect relative to the base?
Experimental study of CG location of symmetrical shape
Hard
A.At a distance from the base
B.At a distance from the base
C.At a distance from the base
D.At a distance from the base
Correct Answer: At a distance from the base
Explanation:
The C.G. of a uniform triangle is its centroid, located one-third of the altitude from the base.
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44For a uniform semicircular lamina of radius , an experiment measures the C.G. on the symmetry axis. Which theoretical distance from the diameter should be used for comparison?
Experimental study of CG location of symmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
A semicircular lamina has its centroid at from the diameter; applies to a semicircular arc.
Incorrect! Try again.
45A uniform annular lamina is suspended from several points. All plumb lines intersect at the center of its empty hole. How should this observation be interpreted?
Experimental study of CG location of symmetrical shape
Hard
A.It proves the annulus has radially varying density
B.It is invalid because a C.G. must lie in material
C.It is valid because the C.G. may lie outside the material
D.It shows that the suspension points were unsuitable
Correct Answer: It is valid because the C.G. may lie outside the material
Explanation:
Rotational symmetry fixes the C.G. at the common center, even though that point lies in the hole.
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46A symmetric plate of mass has its C.G. at the origin. A small mass is attached at . Where should a repeated suspension experiment locate the new C.G.?
Experimental study of CG location of symmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Taking moments about the original C.G. gives .
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47A rectangular lamina occupies and , but its areal density is , where . What C.G. should the suspension experiment reveal?
Experimental study of CG location of symmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The density gradient breaks left-right mass symmetry. Evaluating the first moment gives , while symmetry keeps .
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48A uniform symmetric lamina is first suspended from a point lying exactly on its only symmetry axis. The traced plumb line coincides with that axis. Why is a second suspension from an off-axis point still useful?
Experimental study of CG location of symmetrical shape
Hard
A.It determines the coordinate of the C.G. along the axis
B.It changes the C.G. to a more measurable position
C.It makes the symmetry axis physically vertical forever
D.It proves that the lamina has uniform thickness
Correct Answer: It determines the coordinate of the C.G. along the axis
Explanation:
The first trial only constrains the C.G. to the symmetry axis; an independent plumb line locates its position along that axis.
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49Each traced plumb line has the same small perpendicular-position uncertainty. Which pair of suspension lines generally gives the smallest uncertainty in their intersection?
Experimental study of CG location of symmetrical shape
Hard
A.Lines meeting at approximately
B.Lines meeting at approximately
C.Lines meeting at approximately
D.Lines meeting at approximately
Correct Answer: Lines meeting at approximately
Explanation:
Intersection uncertainty is amplified roughly by , so nearly perpendicular lines provide the best-conditioned estimate.
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50Several suspension lines for a symmetric lamina form a small cluster rather than one exact intersection, while one line misses the cluster widely. What is the best experimental response?
Experimental study of CG location of symmetrical shape
Hard
A.Discard all trials and use only geometric symmetry
B.Select the two lines whose crossing looks most central
C.Repeat the suspect trial and check its suspension setup
D.Average every line without examining the apparatus
Correct Answer: Repeat the suspect trial and check its suspension setup
Explanation:
A widely inconsistent line may result from friction, airflow, or tracing error and should be investigated before inclusion or rejection.
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51An L-shaped uniform lamina is formed from a square by removing the upper-right square. Taking the original square's lower-left corner as the origin, where should suspension lines intersect?
Experimental study of CG location of unsymmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
Treating the removed square as negative area gives .
Incorrect! Try again.
52A uniform plate has a circular hole of radius centered at . Which predicted C.G. should be compared with the plumb-line result?
Experimental study of CG location of unsymmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The hole is negative area, so ; horizontal symmetry keeps .
Incorrect! Try again.
53A uniform triangular lamina has vertices , , and . At which point should three accurately traced suspension lines concur?
Experimental study of CG location of unsymmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The centroid coordinates of a uniform triangle are the averages of the vertex coordinates: .
Incorrect! Try again.
54For a strongly concave uniform lamina, the extensions of the suspension lines intersect in an open region outside the lamina. Which assessment is correct?
Experimental study of CG location of unsymmetrical shape
Hard
A.The result requires the lamina to be nonuniform
B.The result proves that every suspension was unstable
C.The result violates the definition of the C.G.
D.The result can be valid for a concave body
Correct Answer: The result can be valid for a concave body
Explanation:
A body's C.G. need not lie within its material, although it must lie within the convex hull of a positive mass distribution.
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55For an irregular lamina, equal-precision suspension trials produce , , and . Why is simply choosing the intersection of the first two lines inferior to a least-squares estimate?
Experimental study of CG location of unsymmetrical shape
Hard
A.It ignores information supplied by the third measurement
B.It assumes that the lamina has uniform areal density
C.It changes the measured mass of the irregular lamina
D.It requires the first two lines to be perpendicular
Correct Answer: It ignores information supplied by the third measurement
Explanation:
With comparable uncertainties, a least-squares estimate uses all measured constraints rather than arbitrarily privileging two trials.
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56An irregular plate of mass has unknown C.G. coordinate . After a mass is attached at coordinate , suspension trials measure the combined C.G. at . Which expression recovers ?
Experimental study of CG location of unsymmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
From , rearrangement gives the original plate's C.G. coordinate.
Incorrect! Try again.
57In body-fixed coordinates, an irregular lamina is suspended at . The downward plumb-line direction is the unit vector . Which locus correctly represents possible C.G. positions from this trial?
Experimental study of CG location of unsymmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
At stable equilibrium the C.G. lies vertically below the suspension point, so it lies on the downward ray beginning at .
Incorrect! Try again.
58A rectangular lamina consists of a left region with areal density and a right region with density . Using the lower-left corner as the origin, where should the C.G. be measured?
Experimental study of CG location of unsymmetrical shape
Hard
A.
B.
C.
D.
Correct Answer:
Explanation:
The two regions have masses proportional to and , with centers at and , so .
Incorrect! Try again.
59Accurate suspension lines for a large irregular lamina fail to concur, and the discrepancy changes systematically with the lamina's orientation in a region having a strong gravitational gradient. What assumption of the standard method is most directly violated?
Experimental study of CG location of unsymmetrical shape
Hard
A.The plumb bob must have the same mass as the lamina
B.The suspension point must lie directly above the laboratory floor
C.The lamina must have at least one geometric symmetry axis
D.Gravity acts as a uniform parallel field across the lamina
Correct Answer: Gravity acts as a uniform parallel field across the lamina
Explanation:
The usual construction assumes effectively parallel gravitational forces, producing an orientation-independent resultant through one fixed C.G.
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60A uniform irregular lamina is enlarged geometrically by a scale factor while preserving thickness and density. Coordinates are measured from corresponding origins. How should its experimentally determined C.G. coordinates change?
Experimental study of CG location of unsymmetrical shape
Hard
A.They scale by while the mass scales as
B.They scale by while the mass scales as
C.They scale by while the mass scales as
D.They remain unchanged while the mass scales as
Correct Answer: They scale by while the mass scales as
Explanation:
Uniform planar scaling multiplies every centroid coordinate by and every area, hence mass at fixed thickness and density, by .
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