Chapter 3, Problem 93P

(a)

To determine

The amount of force applied to the top block.

Answer to Problem 93P

The amount of force applied to the top block is $9.9\text{N}$ .

Explanation of Solution

Below figure shows force acting on top and bottom block due to rope on top.

Write the expression for net force along vertical of Mass 1.

    $\begin{array}{c}{N}_1-M_{1}g=0\\ N_1=M_{1}g\end{array}$

Here, $N_1$ is the normal force on block 1, $M_1$ is the mass, and $g$ is the acceleration due to gravity.

The normal force $N_1$ for the top block and normal force $N_2$ for the bottom block are action reaction pairs.

    $N_1=N_2$

The top block will slip if it is pulled with more force than the maximum static friction force.

    $F_f={\mu }_{\text{S}}{N}_1$

Here, $F_f$ is the frictional force and ${\mu }_{\text{S}}$ is the coefficient of static friction.

Use $M_{1}g$ for $N_1$ in the above expression and rewrite.

    $F_f={\mu }_{\text{S}}\left(M_{1}g\right)$        (I)

Friction is the only force acting on the block which is not being pulled.

Write the expression for the friction force on the mass not attached to the rope.

    $F_f=M^{\prime }{a}_1$        (II)

Here, $M^{\prime }$ is the mass of the block that is not attached to the rope directly, and $a_1$  is the acceleration.

Equate (I) and (II).

    $\begin{array}{c}{M}^{\prime }a={\mu }_{\text{S}}\left(M_{1}g\right)\\ a=\frac{{\mu }_{\text{S}}\left(M_{1}g\right)}{M^{\prime }}\end{array}$        (III)

Write the expression for the pulling force.

    $F_p=\left(M_1+M_2\right)a$        (IV)

Here, $F_p$ is the pulling force.

Conclusion:

Here, $M_1$ is being pulled, so $M^{\prime }=M_2$

Substitute $\text{3}\text{.0 kg}$ for $M_1$, $\text{9}\text{.8 m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}$ for $g$ , $\text{0}\text{.21}$ for ${\mu }_{\text{S}}$ , and $\text{5}\text{.0 kg}$ for $M_2$ in (III) to find $a_1$.

    $\begin{array}{c}{a}_1=\frac{(\text{0}\text{.21})(\text{3}\text{.0 kg})(\text{9}\text{.8 m}\text{}\text{/}\text{}{\text{s}}^{\text{2}})}{\text{(5}\text{.0 kg)}}\\ =\text{1}\text{.23}\text{m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\end{array}$

Substitute $\text{1}\text{.23}\text{m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}$ for $a_1$ , $\text{3}\text{.0 kg}$ for $M_1$, and $\text{5}\text{.0 kg}$ for $M_2$ in (IV) to find $F_{p1}$.

    $\begin{array}{c}{F}_{p1}=\left(\text{3}\text{.0 kg}+5.\text{0 kg}\right)\left(\text{1}\text{.23}\text{m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\right)\\ =9.9\text{N}\end{array}$

Therefore, the amount of force applied to the top block is $9.9\text{N}$ .

(b)

To determine

The amount of force applied to the bottom block.

Answer to Problem 93P

The amount of force applied to the bottom block is $16\text{N}$ .

Explanation of Solution

Now $M_2$ is pulled. so $M^{\prime }=M_1$

Rewrite equation (III).

    $\begin{array}{c}a=\frac{{\mu }_{\text{S}}\left(M_{1}g\right)}{M_1}\\ ={\mu }_{\text{S}}g\end{array}$        (V)

Conclusion:

Here, $M_1$ is being pulled, so $M^{\prime }=M_2$

Substitute $\text{9}\text{.8 m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}$ for $g$ , $\text{0}\text{.21}$ for ${\mu }_{\text{S}}$ in (V) to find $a_2$.

    $\begin{array}{c}{a}_2=(\text{0}\text{.21})(\text{9}\text{.8 m}\text{}\text{/}\text{}{\text{s}}^{\text{2}})\\ =2.06\text{ m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\end{array}$

Substitute $2.06\text{ m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}$ for $a_2$ , $\text{3}\text{.0 kg}$ for $M_1$, and $\text{5}\text{.0 kg}$ for $M_2$ in (IV) to find $F_{p2}$.

    $\begin{array}{c}{F}_{p2}=\left(\text{3}\text{.0 kg}+5.\text{0 kg}\right)\left(2.06\text{ m}\text{}\text{/}\text{}{\text{s}}^{\text{2}}\right)\\ =16\text{N}\end{array}$

Therefore, the amount of force applied to the bottom block is $16\text{N}$ .