Chapter 4, Problem 72P

(a)

To determine

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

Answer to Problem 72P

The net force on the top link is $6.2\text{N}$ .

Explanation of Solution

Given:

The mass of each link is $m=0.10\text{kg}$ .

The upward acceleration of the chain is $a=2.5\text{m}/{\text{s}}^2$ .

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

  

Figure 1

In the above figure, the force applied on the top link of the chain is $F_{\text{app}}=F_1$ .

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

  $F_1-5\left(mg\right)=5ma$

Calculation:

The net force on the top link is calculated as,

  $\begin{array}{l}{F}_1-5\left(mg\right)=5ma\\ F_1=5ma+5\left(mg\right)\\ F_1=5\left(0.10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2+2.5\text{m}/{\text{s}}^2\right)\\ F_1=6.2\text{N}\end{array}$

Conclusion:

Therefore, the net force on the top link is $6.2\text{N}$ .

(b)

To determine

The net force on each of the link.

Answer to Problem 72P

The net force that acts on each of the link is $0.25\text{N}$ .

Explanation of Solution

Formula used:

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

  $F_{\text{net}}=ma$

Calculation:

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

  $\begin{array}{c}{F}_{\text{net}}=ma\\ =\left(0.10\text{kg}\right)\left(2.5\text{m}/{\text{s}}^2\right)\\ =0.25\text{N}\end{array}$

Conclusion:

Therefore, the net force that acts on each of the link is $0.25\text{N}$ .

(c)

To determine

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

Answer to Problem 72P

The force exerted on the second link is $4.9\text{N}$ , on the third link is $3.7\text{N}$ , on the fourth link is $2.5\text{N}$ and on the last link is $1.2\text{N}$ .

Explanation of Solution

Formula used:

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

  $\begin{array}{l}{F}_1-F_2-mg=ma\\ F_2=F_1-m\left(g+a\right)\end{array}$

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

  $\begin{array}{l}{F}_2-F_3-mg=ma\\ F_3=F_2-m\left(g+a\right)\end{array}$

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

  $F_4=F_3-m\left(g+a\right)$

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

  $F_5=F_4-m\left(g+a\right)$

Calculation:

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

  $\begin{array}{c}{F}_2=F_1-m\left(g+a\right)\\ =6.15\text{N}-\left(0.10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2+2.5\text{m}/{\text{s}}^2\right)\\ =4.9\text{N}\end{array}$

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

  $\begin{array}{c}{F}_3=F_2-m\left(g+a\right)\\ =4.9\text{N}-\left(0.10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2+2.5\text{m}/{\text{s}}^2\right)\\ =3.7\text{N}\end{array}$

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

  $\begin{array}{c}{F}_4=F_3-m\left(g+a\right)\\ =3.68\text{N}-\left(0.10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2+2.5\text{m}/{\text{s}}^2\right)\\ =2.5\text{N}\end{array}$

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

  $\begin{array}{c}{F}_5=F_4-m\left(g+a\right)\\ =2.5\text{N}-\left(0.10\text{kg}\right)\left(9.81\text{m}/{\text{s}}^2+2.5\text{m}/{\text{s}}^2\right)\\ =1.2\text{N}\end{array}$

Conclusion:

Therefore, the force exerted on the second link is $4.9\text{N}$ , on the third link is $3.7\text{N}$ , on the fourth link is $2.5\text{N}$ and on the last link is $1.2\text{N}$ .