Question

In Fig. 6-60, a block weighing22 Nis held at rest against a vertical wall by a horizontal force of magnitude 60 N .The coefficient of static friction between the wall and the block is 0.55 , and the coefficient of kinetic friction between them is 0.38 . In six experiments, a second force is applied to the block and directed parallel to the wall with these magnitudes and directions: (a) 34 N , up, (b) 12 N , up, (c) 48 N , up, (d) 62 N, up, (e) 10 N , down, and (f) 18 N, down. In each experiment, what is the magnitude of the frictional force on the block? In which does the block move (g) up the wall and (h) down the wall? (i) In which is the frictional force directed down the wall?

Short answer

1. Magnitude of frictional force for 12 N down
2. Magnitude of frictional force for 10 N up
3. Magnitude of frictional force for 26 N down
4. Magnitude of frictional force for 23 N down
5. Magnitude of frictional force for 32 N up
6. Magnitude of frictional force for 23 N up
7. The block moves up the wall in case (d) where a > 0 .
8. The block moves down the wall in case (f) where a < 0 .
9. The frictional force $\overrightarrow{f_s}$is directed down in cases (a), (c) and (d).

Step-by-step solution

Given

Horizontal force,$F=60N$.

Coefficient of static friction,${\mu }_s=0.55$.

Coefficient of kinetic friction, ${\mu }_k=0.38$.

Understanding the concept

After reading the problem, analysis of forces in the horizontal direction (where there can be no acceleration) leads to the conclusion that$\mathit{F}\mathbf{=}{\mathit{F}}_{\mathbf{N}}$ , the magnitude of the normal force is 60 N . The maximum possible static friction force is therefore ${\mathit{\mu }}_{\mathbf{s}}{\mathit{F}}_{\mathbf{N}}\mathbf{=}\mathbf{33}\mathbf{}\mathit{N}$, and the kinetic friction force (when applicable) is ${\mathit{\mu }}_{\mathbf{k}}\mathbf{}{\mathit{F}}_{\mathbf{N}}\mathbf{=}\mathbf{23}\mathbf{}\mathit{N}$. Further, using Newton’s second law, we can solve the given problem.

 Calculate the magnitude of the frictional force on the block when the second force is , up 34 N, up

(a)

In this case, $\overrightarrow{P}=34\mathrm{N}$upward. Assuming $\overrightarrow{f}$ points down, then Newton's second law for the $y$ leads to

$P-mg-f=ma$

if we assume $f=f_s$and $a=0$, we obtain

$$\begin{gathered} f=\left(34-22\right)\mathrm{N} \\ =12\mathrm{N} \end{gathered}$$.

This is less than$f_{s,max}$, which shows the consistency of our assumption.

The answer is:

$\overrightarrow{f_s}=12\mathrm{N}$ Down.

Calculate the magnitude of the frictional force on the block when the second force is 12 N , up

(b)

In this case, $\overrightarrow{P}=12\mathrm{N}$upward. The above equation, with the same assumptions as in part (a), leads to

$$\begin{gathered} f=\left(12-22\right)\mathrm{N} \\ =-10\mathrm{N} \end{gathered}$$ .

Thus, $\left|f_s\right|<f_{s,max}$, justifying our assumption that the block is stationary, but its negative value tells us that our initial assumption about the direction of $\overrightarrow{f}$is incorrect in this case. Thus, the answer is:

$\overrightarrow{f_s}=10\mathrm{N}$up.

Calculate the magnitude of the frictional force on the block when the second force is 48 N , up

(c) In this case,$\overrightarrow{P}=48\mathrm{N}$upward. The above equation, with the same assumptions as in part (a), leads to

$$\begin{gathered} f=\left(48-22\right)\mathrm{N} \\ =26\mathrm{N} \end{gathered}$$.

Thus, we again have $f_s<f_s$, max, and our answer is:

$\overrightarrow{f_s}=26\mathrm{N}$down.

Calculate the magnitude of the frictional force on the block when the second force is 62 N , up

(d)

In this case, $\overrightarrow{P}=62\mathrm{N}$ upward. The above equation, with the same assumptions as in part (a), leads to

$$\begin{gathered} f=\left(62-22\right)\mathrm{N} \\ =40\mathrm{N} \end{gathered}$$ ,

Which is larger than $f_s$, max, $-$ invalidating our assumptions. Therefore, we take $f=f_k$and $a\ne 0$ in the above equation; if we wished to find the value of a we would find it to be positive, as we should expect.

The answer is:

$\overrightarrow{f_k}=23\mathrm{N}$down.

Calculate the magnitude of the frictional force on the block when the second force is 10 N , down

(e)

In this case, $\overrightarrow{P}=10\mathrm{N}$downward. The above equation (but with $P$replaced with $-P$) with the same assumptions as in part (a), leads to

$$\begin{gathered} f=\left(-10-22\right)\mathrm{N} \\ =-32\mathrm{N} \end{gathered}$$.

Thus, we have$\left|f_s\right|<f_{s,max}$, justifying our assumption that the block is stationary, but its negative value tells us that our initial assumption about the direction of$\overrightarrow{f}$is incorrect in this case.

Thus, the answer is:

$\overrightarrow{f_s}=32\mathrm{N}$up.

 Calculate the magnitude of the frictional force on the block when the second force is 18 N , down

(f)

In this case, $\overrightarrow{P}=18\mathrm{N}$ downward. The above equation (but with $P$replaced with $-P$) with the same assumptions as in part (a), leads to

$$\begin{gathered} f=\left(18-22\right)\mathrm{N} \\ =-40\mathrm{N} \end{gathered}$$,

Which is larger (in absolute value) than$f_{s,max}$,$-$invalidating our assumptions. Therefore, we take$f=f_k$and$a\ne 0$in the above equation; if we wished to find the value of$a$we would find it to be negative, as we should expect. The answer is:

$\overrightarrow{f_k}=23\mathrm{N}$ Up.

 Figure out if in which case does the block move up the wall

(g)

The block moves up the wall in case (d) where a > 0 .

Figure out if in which case does the block move down the wall

(h)

The block moves down the wall in case (f) where a < 0 .

Figure out if in which case is the frictional force directed down the wall

(i)

The frictional force $\overrightarrow{f_s}$is directed down in cases (a), (c) and (d).