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Two uniform, 75.0 g marbles 2.00 c³¾ in diameter are stacked as shown in given figure in a container that is 32.00 c³¾wide. (a) Find the force that the container exerts on the marbles at the points of contact A, B, and C. (b) What force does each marble exert on the other?

Short Answer

Expert verified
  1. FA=FC=0.42 N and FB=1.47 N
  2. The force that each marble exerts on the other is 0.84 N

Step by step solution

01

Force exert at the point

The force a charge exerts on another point is independent of the existence of other charges, according to the principle of superposition, and the net force a charge experiences as a result of the number of other charges acting on it is the vector sum of those forces.

02

Identification of given data

Mass of both marbles arem=75.0 g=0.075 k²µ

The diameter of marbles is d=2.00 c³¾=0.02″¾

Wide container. So, L=32.00 c³¾

03

Finding the force that the container exerts on the marbles at the points of contact A, B, and C

(a)

At this pointAand C, the forces have the same magnitude even if they are in the opposite direction

So, FA=FC

At the center of the bottom marble P, the summation of torque is zero.

So,

∑τP=0FA(³¦´Ç²õθ)+FC(³¦´Ç²õθ)−W=0FC(2³¦´Ç²õθ)−W=0

FC=mg2cos30° (1)

Now, put numerical values in equation (1), we get

FC=(0.0075 k²µ)(9.8″¾/s2)2cos30°=0.42 N

So, FA=FC=0.42 N

Now, at point B, the exerted force is vertically due to the weight of the two marbles, so, it is given by

FB=2W=2(0.0075 k²µ)(9.8″¾/s2)=1.47 N

So, FB=1.47 N

Hence, FA=FC=0.42 N and FB=1.47 N

04

Finding the force that each marble exerts on the other.

(b)

Each marble exerts a normal forceon the other and this normal force makes an angle θ=30°with the vertical axis.

So, the normal force in relation to horizontal forceFAcould be given by

FA=Nsin30°N=FAsin30°=0.42 N0.5=0.84 N

Hence, the force that each marble exerts on the other is 0.84 N.

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