Physics for Scientists and Engineers, Technology Update (No access codes included)
Physics for Scientists and Engineers, Technology Update (No access codes included)
9th Edition
ISBN: 9781305116399
Author: Raymond A. Serway, John W. Jewett
Publisher: Cengage Learning
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Chapter 5, Problem 5.79AP

Two blocks of masses m1 and m2, are placed on a table in contact with each other as discussed in Example 5.7 and shown in Figure 5.13a. The coefficient of kinetic friction between the block of mass m1 and the table is μ1, and that between the block of mass m2 and the table is μ2. A horizontal force of magnitude F is applied to the block of mass m1. We wish to find P, the magnitude of the contact force between the blocks. (a) Draw diagrams showing the forces for each block. (b) What is the net force on the system of two blocks? (c) What is the net force acting on m1? (d) What is the net force acting on m2? (e) Write Newton’s second law in the x direction for each block. (f) Solve the two equations in two unknowns for the acceleration a of the blocks in terms of the masses, the applied force F, the coefficients of friction, and g. (g) Find the magnitude P of the contact force between the blocks in terms of the same quantities.

Chapter 5, Problem 5.79AP, Two blocks of masses m1 and m2, are placed on a table in contact with each other as discussed in

(a)

Expert Solution
Check Mark
To determine

The free body diagram of each block with forces.

The free body diagram of an object represents the direction and magnitude of forces acting on the body.

Explanation of Solution

The mass of block 1 is m1, the mass of block 2 is m2, the coefficient of kinetic friction between m1 and the table is μ1, the coefficient of kinetic friction between m2 and the table is μ2 and the magnitude of horizontal force is F.

The free body diagram of the book is given below.

Physics for Scientists and Engineers, Technology Update (No access codes included), Chapter 5, Problem 5.79AP

Figure (1)

The sum of all vertical forces is zero because the block moves on a horizontal surface. So the vertical acceleration, ay=0.

Write the net force in the y-direction for mass m1 using Newton’s law

    F1y=m1aym1g+n1=0n1=m1g

Here, F1y is the vertical force of block 1, ay is the acceleration in the y-direction, n1 is the normal force of block 1 and g is the acceleration due to gravity.

Write the net force in the y-direction for mass m2 using Newton’s law

    F2y=m2aym2g+n2=0n2=m2g

Here, F2y is the vertical force of block 2 and n2 is the normal force of block 2.

Write the equation for kinetic friction for block 1

    f1=μ1n1=μ1m1g

Here, f1 is the frictional force due to mass m1.

Write the equation for kinetic friction for block 2

    f2=μ2n2=μ2m2g

Here, f2 is the frictional force due to mass m2.

In the figure, P is the contact force that arises due to the contact between m1 and m2.

Conclusion:

Therefore, the free body diagram of each block to show the forces is given in figure I.

(b)

Expert Solution
Check Mark
To determine

The net force on the system of two blocks.

Answer to Problem 5.79AP

The net force on the system of two blocks is the external force applied minus the frictional force.

Explanation of Solution

Write the expression for the net force in x-direction for the system of two blocks from the figure I,

  F=Ff1f2+PP=Ff1f2

Here, F is the net force on the system of two blocks.

Conclusion:

Therefore, the net force on the system of two blocks is the external force applied minus the frictional force.

(c)

Expert Solution
Check Mark
To determine

The net force acting on m1.

Answer to Problem 5.79AP

The net force acting on m1 is Ff1P.

Explanation of Solution

Write the expression for the net force in x-direction for the system of two blocks from the figure I,

  F1x=Ff1P

Here, F1x is the net force on the block of mass m1 .

Conclusion:

Therefore, the net force acting on m1 is Ff1P.

(d)

Expert Solution
Check Mark
To determine

The net force acting on m2.

Answer to Problem 5.79AP

The net force acting on m2 is Pf2.

Explanation of Solution

Write the expression for the net force in x-direction for the system of two blocks from the figure I,

  F2x=f2+P=Pf2

Here, F2x is the net force on the block of mass m2 .

Conclusion:

Therefore, the net force acting on m2 is Pf2.

(e)

Expert Solution
Check Mark
To determine

The Newton’s second law in the x direction for each block.

Answer to Problem 5.79AP

The Newton’s second law in the x direction, for m1 is FP=m1a and for m2 is P=m2a.

Explanation of Solution

The block has on a horizontal acceleration ax=a.

Write the Newton’s second law for block 1

  F1x=m1ax

Substitute Ff1P for F1x in the above equation

  Ff1P=m1a

Substitute μ1m1g for f1 in the above equation

    Fμ1m1gP=m1a

Write the Newton’s second law for block 2

  F2x=m2ax

Substitute Pf2 for F2x in the above equation

  Pf2=m2a

Substitute μ2m2g for f2 in the above equation

    Pμ2m2g=m2a

Conclusion:

Therefore, the Newton’s second law in the x direction, for m1 is Fμ1m1gP=m1a and for m2 is Pμ2m2g=m2a.

(f)

Expert Solution
Check Mark
To determine

The acceleration of the blocks.

Answer to Problem 5.79AP

The acceleration of the blocks is (Fμ1m1gμ2m2gm1+m2).

Explanation of Solution

Write the Newton’s second law is for block 1

    FPμ1m1g=m1a                                                              (I)

Write the Newton’s second law is for block 2

    Pμ2m2g=m2a                                                                 (II)

Conclusion:

Add the equation (I) and equation (II) and solve for a.

    FPμ1m1g+Pμ2m2g=m1a+m2aa=(Fμ1m1gμ2m2gm1+m2)

Therefore, the acceleration of the blocks is (Fμ1m1gμ2m2gm1+m2).

(g)

Expert Solution
Check Mark
To determine

The magnitude of the contact force between the blocks in terms of acceleration, mass, applied force and the friction coefficient.

Answer to Problem 5.79AP

The magnitude P of the contact force between the blocks is (m2m1+m2)(F+(μ2μ1)m1g).

Explanation of Solution

Recall the equation (II).

    Pμ2m2g=m2aP=μ2m2g+m2a

Substitute (Fμ1m1gμ2m2gm1+m2) for a in above expression.

    P=μ2m2g+m2(Fμ1m1gμ2m2gm1+m2)=(m2m1+m2)(F+(μ2μ1)m1g)

Conclusion:

Therefore, the magnitude P of the contact force between the blocks is (m2m1+m2)(F+(μ2μ1)m1g).

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Chapter 5 Solutions

Physics for Scientists and Engineers, Technology Update (No access codes included)

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