Question

A $\mathbf{12}\mathbf{}\mathbf{N}$horizontal force $\stackrel{\mathbf{\rightharpoonup }}{\mathbf{F}}$pushes a block weighing $\mathbf{5}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{N}$against a vertical wall (Fig. 6-26). The coefficient of static friction between the wall and the block is $\mathbf{0}\mathbf{.}\mathbf{60}$, and the coefficient of kinetic friction is $\mathbf{0}\mathbf{.}\mathbf{40}$. Assume that the block is not moving initially. (a) Will the block move? (b) In unit-vector notation, what is the force on the block from the wall?

Short answer

(a) Block will not slide

(b) The force on the block from the wall is, ${}^{F_{BW}=(-12)\hat{i}+5\hat{j}}$

Step-by-step solution

Given

Force on the block,${}^{F=12N}$

Weight of the block,${}^{W=5.0N}$

Coefficient of static friction, ${}^{{\mu }_s=0.60}$

Coefficient of kinetic friction, ${}^{{\mu }_k=0.40}$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it.Use the Newton's 2nd law of motion along vertical and horizontal direction to check if the block will move or not. According to Newton's 2nd law of motion, a force applied to an object at rest causes it to accelerate in the direction of the force.

Formula:

$F_{net}=\sum ma$

where, F is the net force, m is mass and a is an acceleration.

Determining the free body diagram

(a) Determining if block will slide or not

By using Newton’s 2nd law of motion along horizontal direction,

$$\begin{gathered} \mathrm{N}=\mathrm{F} \\ =12\mathrm{N} \end{gathered}$$

then, the maximum static frictional force of the block is,

$$\begin{gathered} f_{ma{x}_s}={\mu }_{s}N \\ =(0.60)\left(12\right) \\ =7.2N \end{gathered}$$

Therefore, as the $f_{ma{x}_s}>W$, block will not slide.

(b) Determining the force on the block from the wall

Since, the block does not move then force on the block is only ${}^{f_s}$and normal force from the wall is,

$$\begin{gathered} F_{BW}=F_N\hat{i}+f_s\hat{j} \\ =(-12)\hat{i}+5\hat{j} \end{gathered}$$

Hence, the force on the block from the wall is${}^{F_{BW}=(-12)\hat{i}+5\hat{j}}$