Physics for Scientists and Engineers
Physics for Scientists and Engineers
6th Edition
ISBN: 9781429281843
Author: Tipler
Publisher: MAC HIGHER
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Chapter 4, Problem 72P

(a)

To determine

The magnitude of the force F exerted on the top link by hand.

(a)

Expert Solution
Check Mark

Answer to Problem 72P

The net force on the top link is 6.2N .

Explanation of Solution

Given:

The mass of each link is m=0.10kg .

The upward acceleration of the chain is a=2.5m/s2 .

Formula used:

The diagram for the five chains linked to each other and the force that is exerted on the link one and two is shown in Figure 1

  Physics for Scientists and Engineers, Chapter 4, Problem 72P

Figure 1

In the above figure, the force applied on the top link of the chain is Fapp=F1 .

The expression for the net force exerted on the top link is given by,

  F15(mg)=5ma

Calculation:

The net force on the top link is calculated as,

  F15(mg)=5maF1=5ma+5(mg)F1=5(0.10kg)(9.81m/ s 2+2.5m/ s 2)F1=6.2N

Conclusion:

Therefore, the net force on the top link is 6.2N .

(b)

To determine

The net force on each of the link.

(b)

Expert Solution
Check Mark

Answer to Problem 72P

The net force that acts on each of the link is 0.25N .

Explanation of Solution

Formula used:

The expression for the net force on each of the link is given by,

  Fnet=ma

Calculation:

The net force on each of the link is given by,

  Fnet=ma=(0.10kg)(2.5m/ s 2)=0.25N

Conclusion:

Therefore, the net force that acts on each of the link is 0.25N .

(c)

To determine

The magnitude of the force that each link exerts on the link below it.

(c)

Expert Solution
Check Mark

Answer to Problem 72P

The force exerted on the second link is 4.9N , on the third link is 3.7N , on the fourth link is 2.5N and on the last link is 1.2N .

Explanation of Solution

Formula used:

The expression for the net force exerted by the top link on the link below it is given by,

  F1F2mg=maF2=F1m(g+a)

The expression for the net for exerted by the second link on the third link is given by,

  F2F3mg=maF3=F2m(g+a)

The expression for the force exerted by the third link on the fourth link is given by,

  F4=F3m(g+a)

The expression for the force exerted by the fourth link on the last link is given by,

  F5=F4m(g+a)

Calculation:

The force exerted by the first link on the second link is calculated as,

  F2=F1m(g+a)=6.15N(0.10kg)(9.81m/ s 2+2.5m/ s 2)=4.9N

The force exerted by the second link on the third link is calculated as,

  F3=F2m(g+a)=4.9N(0.10kg)(9.81m/ s 2+2.5m/ s 2)=3.7N

The force exerted by the third link on the fourth link is calculated as,

  F4=F3m(g+a)=3.68N(0.10kg)(9.81m/ s 2+2.5m/ s 2)=2.5N

The force exerted by the fourth link on the last link is calculated as,

  F5=F4m(g+a)=2.5N(0.10kg)(9.81m/ s 2+2.5m/ s 2)=1.2N

Conclusion:

Therefore, the force exerted on the second link is 4.9N , on the third link is 3.7N , on the fourth link is 2.5N and on the last link is 1.2N .

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