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Fundamentals Of Physics
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Short Answer

In three experiments, three different horizontal forces are applied to the same block lying on the same countertop. The force magnitudes are F1=12N,F2=8N, F3=4N. In each experiment, the block remains stationary in spite of the applied force. Rank the forces according to (a) the magnitude of the static frictional force on the block from the countertop and (b) the maximum value role="math" localid="1660904123305" fs,max of that force, greatest first.

a) The magnitude of the static frictional force on the block from the countertop is given by fs1>fs2>fs3.

b) The maximum values fsmaxof those forces are given by fs1max=fs2max=fs3max .

See the step by step solution

Step by Step Solution

Step 1: The given data

The magnitudes of the force are: F1=12N, F2=8N, and F3=4N.

Step 2: Understanding the concept of the force

We have to use Newton’s 2nd law of motion along with x and y directions on each block. Now, using the given data of the magnitudes of the force, we can get the behavior of the frictional force and the maximum frictional force.

Formulae:

The force according to Newton’s second law,

F=ma (1)

Step 3: a) Calculation of the magnitude of the static frictional force

Free body diagram of blocks:

Blocks are stationary therefore the acceleration of the blocks along x is 0.

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the first block can be given using equation (1) as:

(Case-1)

Fx=maxF1-fs1=0 fs1=F1 fs1=12 N

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the first block can be given using equation (1) as:

(Case-2)

Fx=maxF2=fs2=0 fs2=F2 fs2=8 N

By using Newton’s 2nd law along the x-direction, the net of the horizontal force for the third block can be given using equation (1) as:

(Case-3)

Fx=maxF3=fs3=0 fs3=F3 fs3=4 N

So, the block with F1=12 N force has the greatest frictional force fs1.

Hence, the magnitude of the forces is ordered as fs1>fs2>fs3.

Step 4: b) Calculation of the magnitude of the maximum force

The value of the magnitude fsmaxof the static frictional force is defined as,

fsmax=μN

So, fsmax is directly proportional to the normal force N.

By using Newton’s 2nd law along the vertical direction, the net vertical forces acting on the body can be given using equation (1) as (since blocks are stationary ay=0.)

Fy=mayN-Mg=0 N=Mg

Here, the mass of the blocks (M) is the same, and g is constant so normal force also remains the same in all blocks.

Therefore, fsmax is the same for all blocks that imply the magnitude values as fs1max=fs2max=fs3max.

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