Chapter 4, Problem 46P

To determine

The mass of the box which is held in place against a rough vertical wall by using a force directed upwards at an angle.

Answer to Problem 46P

Solution:

The mass of the box is found to be 1.9 kg.

Explanation:

The box of mass m is held in place when the net force on the box is zero. It starts to slide when the force acting downwards is more than the force acting in the upward direction.

Draw the free body diagram of the box.

The box is pushed against the wall by a force F acting at an angle $\theta$ to the horizontal. The weight $mg$ of the box acts downwards. Since the box tends to slide down, the force of friction $F_{fr}$ acts in the upward direction and the normal force $F_N$ acts perpendicular to the wall.

Resolve the force F into two components $F\sin \theta \text{ and }F\cos \theta$, as shown in the figure.

For horizontal equilibrium,

    $F\cos \theta =F_N$…………(1)

For vertical equilibrium,

    $F_{fr}+F\sin \theta =mg$…………(2)

The force of friction $F_{fr}$ and normal force $F_N$ are related as follows.

    $F_{fr}={\mu }_s{F}_N$…………(3)

Substitute equations (1)and (3)in equation (2).

${\mu }_{s}F\cos \theta +F\sin \theta =mg$

Simplify and write the expression for m.

    $m=\frac{F\left({\mu }_s\cos \theta +\sin \theta \right)}{g}$

Explanation of Solution

Given:

The coefficient of static friction ${\mu }_s=0.40$

The coefficient of kinetic friction ${\mu }_k=0.30$

The magnitude of the applied force $F=23\text{ N}$

The angle made by the force to the horizontal $\theta =28°$

Formula used:

    $m=\frac{F\left({\mu }_s\cos \theta +\sin \theta \right)}{g}$

Calculation:

Substitute the given values of F, ${\mu }_s$, ${\mu }_k$, $\theta$ and 9.8 m/s² for g in the formula used.

    $\begin{array}{c}m=\frac{F\left({\mu }_s\cos \theta +\sin \theta \right)}{g}\end{array}$

        $\begin{array}{c}=\frac{\left(23\text{ N}\right)\left[\left(0.40\right)\left(\cos 28°\right)+\left(\sin 28°\right)\right]}{9.8{\text{ m/s}}^{-2}}\end{array}$

        $\begin{array}{c}=1.93\text{ kg}\end{array}$