91Ó°ÊÓ

In such a problem, first, the free body diagram of the box should be drawn. When the hanging weight is smaller than the box’s weight, the object does not move.

Also, the tension force will be equivalent to the weight of the hanging body.

Given data:

The weight of the box is $W=77\mathrm{N}$.

The weights hanging on the other side of the pulley for parts (a), (b), and (c) are $F_T=30\mathrm{N}$, $F_T=60\mathrm{N}$, and $F_T=90\mathrm{N}$, respectively.

The free body diagram is as follows:

The relation to calculate the vertical forces is given by:

role="math" localid="1645422070664" $\begin{array}{c}\mathrm{Î\pounds }{F}_v=0\\ F_N+F_T=W\\ F_N=W-F_T……\left(I\right)\end{array}$

Here, $F_N$is the required force that the table exerts.

On plugging the values in equation (I), you get:

$\begin{array}{c}{F}_N=\left(77\mathrm{N}\right)-\left(30\mathrm{N}\right)\\ F_N=47\mathrm{N}\end{array}$

Thus, $F_N=47\mathrm{N}$is the required force.

In order to calculate the force exerted by the table for part (b), substitute the values in equation (I).

$\begin{array}{l}{F}_N=\left(77\mathrm{N}\right)-\left(60\mathrm{N}\right)\\ F_N=17\mathrm{N}\end{array}$

Thus, $F_N=17\mathrm{N}$is the required force.

In this case, the hanging weight $F_T=90\mathrm{N}$is more than the weight of the box. Due to this, the box will be lifted upward, and the value of the normal force will be equivalent to zero.

Thus, $F_N=0\mathrm{N}$is the required force.