An 8.0 kgblock of steel is at rest on a horizontal table. The coefficient of static friction between the block and the table is 0.450. A force is to be applied to the block. To three significant figures, what is the magnitude of that applied force if it puts the block on the verge of sliding when the force is directed

(a) horizontally,

(b) upward at$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$from the horizontal, and

(c) downward at$\mathbf{60}\mathbf{.}\mathbf{0}\mathbf{°}$from the horizontal?

• The mass of the block is 8 kg .
• The coefficient of static friction,${\mu }_k=0.45$.

1. The force is horizontal.
2. The force in inclined$60°$upward.
3. The force in inclined $60°$downward.

The problem deals with Newton’s second law of motion, which states that the acceleration of an object is dependent upon the net force acting upon the object and the mass of the object.Use Newton’s second law of motion with zero acceleration to solve this question.

Formula:

$F=ma$

The normal force on the block with horizontal force:

$F_N=mg$

On the verge of sliding, the applied force is equal to the maximum static frictional force:

$f_{s.m}={\mu }_{k}mg$

Substitute the values in the above expression, and we get,

$$\begin{gathered} f_{s.m}=0.45\times \left(8\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ {f}_{s.m}=35.3\mathrm{N} \end{gathered}$$

Thus, the magnitude of applied force is 35.3 N.

The normal force on the block with upward inclined force:

$F_N=mg-F\sin \theta$

Newton’s second law of motion when the block is on verge of sliding:

role="math" localid="1660965590637" $$\begin{gathered} F\cos \theta -f_{s.m}=0 \\ F\cos \theta -{\mu }_k\left(mg-F\sin \theta \right)=0 \end{gathered}$$

Substitute the values in the above expression, and we get,

role="math" localid="1660965905081" $$\begin{gathered} F\cos 60°-\left(0.45\right)\times \left(\left(8\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\mathrm{F}\sin \left(60°\right)\right)=0 \\ \mathrm{F}=39.7\mathrm{N} \end{gathered}$$

Thus, the magnitude of applied force is 39.7 N.

The normal force on the block with upward inclined force:

$F_N=mg+F\sin \theta$

Newton’s second law of motion when the block is on verge of sliding:

$$\begin{gathered} F\cos \theta -f_{s.m}=0 \\ F\cos \theta -{\mu }_k\left(mg+F\sin \theta \right)=0 \end{gathered}$$

Substitute the values in the above expression, and we get,

$$\begin{gathered} F\cos 60°-\left(0.45\right)\times \left(\left(8\mathrm{kg}\right)\times \left(9.8\mathrm{m}/{\mathrm{s}}^2\right)-\mathrm{F}\sin \left(60°\right)\right)=0 \\ \mathrm{F}=320\mathrm{N} \end{gathered}$$

Thus, the magnitude of applied force is 320 N.