Question

Block A in Fig. P5.79 weighs 1.20 N, and block B weighs 3.60 N. The coefficient of kinetic friction between all surfaces is0.300. Find the magnitude of the horizontal force$\overrightarrow{\mathbf{f}}$necessary to drag block B to the left at constant speed (a) if A rests on B and moves with it (Fig. P5.79a), (b) if A is held at rest (Fig. P5.79b).

Short answer

(a) The magnitude of the horizontal force if A rests on B and moves with it is 1.44 N .

(b) The magnitude of the horizontal force if A is held at rest is 1.8 N .

Step-by-step solution

Identification of the given data:

The given data can be listed below as:

• The weight of the block A is ${\mathrm{w}}_1=1.20\mathrm{N}$.
• The weight of the block B is ${\mathrm{w}}_2=3.60\mathrm{N}$.
• The coefficient of the kinetic friction between the surfaces is $\mu =0.300$.

Significance of the frictional force

The friction force mainly resists the motion between two surfaces. The friction force is equal to the product of the frictional coefficient and the normal force.

(a) Determination of the horizontal force in the first case

As block A rests on block B and moves with it, then the normal force will be their added weight.

The equation of the total frictional force is expressed as:

${\mathrm{F}}_1=\mathrm{\mu }\left({\mathrm{w}}_1+{\mathrm{w}}_2\right)$

Here, ${\mathrm{F}}_1$is the frictional force, $\mu$is the coefficient of the kinetic friction between the surfaces, ${\mathrm{w}}_1$ is the weight of the block A and ${\mathrm{w}}_2$is the weight of the block B.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_1=0.3\left(1.20\mathrm{N}+3.60\mathrm{N}\right) \\ =0.3(4.8\mathrm{N}) \\ =1.44\mathrm{N} \end{gathered}$$

The frictional force is required to put the blocks into uniform motion. Hence, it acts like the horizontal force.

Thus, the magnitude of the horizontal force if A rests on B and moves with it is 1.44 N .

(b) Determination of the horizontal force in the second case:

If the block A is held at rest, then the frictional force will act on the lower and also at the upper surface of the block B.

The equation of the frictional force on the upper surface of the block B is expressed as:

${\mathrm{F}}_2={\mathrm{\mu W}}_1$

Here, ${\mathrm{F}}_2$is the frictional force on the upper surface of the block B.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_2=0.3\times 1.2\mathrm{N} \\ =0.36\mathrm{N} \end{gathered}$$

The equation of the frictional force on the lower surface of the block B is expressed as:

role="math" localid="1663771988100" ${\mathrm{F}}_3={\mathrm{\mu W}}_2$

Here, ${\mathrm{F}}_2$is the frictional force on the lower surface of the block B.

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_3=0.3\times 4.8\mathrm{N} \\ =1.44\mathrm{N} \end{gathered}$$

The equation of the total frictional force is expressed as:

${\mathrm{F}}_4={\mathrm{F}}_2+{\mathrm{F}}_3$

Substitute the values in the above equation.

$$\begin{gathered} {\mathrm{F}}_4=0.36\mathrm{N}+1.44\mathrm{N} \\ =1.8\mathrm{N} \end{gathered}$$

The frictional force is the horizontal force required to move the blocks.

The magnitude of the horizontal force necessary to drag block B if A is held at rest is 1.8 N .