Question

A box of mass 40 kghangs motionless from two ropes, as shown in Figure. The angle is $\mathbf{38}\mathbf{°}$. Choose the box as the system. The xaxis runs to the right, the yaxis runs up, and the zaxis is out of the page.

(a) Draw a free-body diagram for the box.

(b) Is$\mathit{d}\overrightarrow{\mathbf{p}}\mathbf{/}\mathit{d}\mathit{t}$of the box zero or nonzero?

(c) What is the ycomponent of the gravitational force acting on the block? (A component can be positive or negative).

(d) What is theycomponent of the force on the block due to rope 2?

(e) What is the magnitude of localid="1657085603204" ${\overrightarrow{\mathbf{F}}}_{\mathbf{2}}$?

(f) What is thexcomponent of the force on the block due to rope 2?

(g) What is the xcomponent of the force on the block due to rope 1?

Short answer

(a)Free body diagram for the block rope system.

(b)The value of$\frac{d\overrightarrow{p}}{dt}$is zero.

(c)The y component of the gravitational force acting on the block is $-392\mathrm{J}$.

(d)The y component of the force on the block due to rope 2 is 306N .

(e)The magnitude of${\overrightarrow{F}}_2$ is 497.4 N .

(f)The x-component of the force on the block due to rope 2 is 306.2 N .

(g)The x-component of the force on the block due to rope 1 is 306.2 N .

Step-by-step solution

Given

A box of mass 40 kg hangs motionless from two ropes, as shown in Figure. The angle is$38°$

Free body diagram of the block rope system.

Tension is acting upwards for which weight counter is balanced by the system downwards.

A box of mass mhangs motionless from two ropes as shown in the following figure.

The free body diagram for the block rope system is shown in the following figure.

Here, ${\overrightarrow{F}}_{grav}$is the gravitational force acting on the block, ${\overrightarrow{F}}_T$is the tension force acting on the first rope, ${\overrightarrow{F}}_{T2}$is the tension force on the second rope.

Calculating the rate of change of momentum of the box.

The velocity of the box is not changing, because it is hanging motionless from two ropes. So, the rate of change of momentum of the box is zero.

$$\begin{gathered} \frac{d\overrightarrow{p}}{dt}=\frac{d\left(m\overrightarrow{v}\right)}{dt} \\ \frac{d\overrightarrow{p}}{dt}=m\frac{d\left(\overrightarrow{v}\right)}{dt} \\ \frac{d\overrightarrow{p}}{dt}=0 \end{gathered}$$

Therefore, the value of $\frac{d\overrightarrow{p}}{dt}$is zero.

Calculating the force due to gravitational force on the box.

The force due to gravitational force (Earth) on the block is pointed to the downward (negative y direction). Therefore, the y-component of the force due to gravitational force (Earth) on the block is as follows.

$F_{gravy}=mg$

Here,m is the mass of the block and gis the acceleration due to gravity.

Substitute 40 kg for mand$-9.8\mathrm{m}/{\mathrm{s}}^2$for g .

$$\begin{gathered} F_{grav,y}=\left(40\mathrm{kg}\right)\left(-9.8\mathrm{m}/{\mathrm{s}}^2\right) \\ {F}_{grav,y}=-392\mathrm{J} \end{gathered}$$

Hence, the y component of the gravitational force acting on the block is $-392\mathrm{J}$.

Calculating the force in the opposite direction.

If the block is in rest position, the y-component of the force on the block due to rope 2 is equal to the force on the block due to gravitational force, but opposite in direction.

${\overrightarrow{F}}_{T2,y}=-{\overrightarrow{F}}_{grav}$

Substitute$-392\mathrm{J}$for${\overrightarrow{F}}_{grav}$.

$$\begin{gathered} {\overrightarrow{F}}_{T2,y}=-\left(-392\mathrm{N}\right) \\ {\overrightarrow{F}}_{T2,y}=392\mathrm{N} \end{gathered}$$

Hence, they component of the force on the block due to rope 2 is 392N .

Balancing the force.

From the above figure, they -component of the force on the block due to rope 2 is balanced with the force on the block due to gravitational force.

$$\begin{gathered} {\overrightarrow{F}}_{T2,y}=mg{F}_{T2}\cos \theta \\ {\overrightarrow{F}}_{T2,y}=mg \end{gathered}$$

Here,$F_{T2}\cos \theta$is the y component of the tension force on the rope 2 .

Rearrange the equation for$F_{T2}$.

$F_{T2}=\frac{mg}{\cos \theta }$

Substitute 40 kg for $m,38°$ for $\theta$, and localid="1657086939393" $9.8\mathrm{m}/{\mathrm{s}}^2$for $g$.

$$\begin{gathered} F_{T2}=\frac{\left(40\mathrm{kg}\right)\left(9.8\mathrm{m}/{\mathrm{s}}^2\right)}{\cos 38°} \\ {F}_{T2}=497.4\mathrm{N} \end{gathered}$$

Therefore, the magnitude of ${\overrightarrow{F}}_2$is 497.4 N .

Calculating the x -component of the force on the block.

The x-component of the force on the block due to rope 2 is,

${\overrightarrow{F}}_{T2,x}=F_{T2}\sin \theta$

Substitute 497.4 N for $F_{T2}$and $38°$for $\theta$.

$$\begin{gathered} {\overrightarrow{F}}_{T2,x}=\left(497.46\mathrm{N}\right)\sin 38° \\ =306.2\mathrm{N} \end{gathered}$$

Hence, the -component of the force on the block due to rope 2 is $306.2\mathrm{N}$.

Balancing the force

The x -component of the force on the block due to rope 1 is balanced with the x -component of the force on the block due to rope 2.

${\overrightarrow{F}}_{T1,x}=F_{T2}\sin \theta$

Substitute 497.4N for $F_{T2}$and $38°$for $\theta$.

$$\begin{gathered} {\overrightarrow{F}}_{T2,x}=\left(497.46\right)\sin 38° \\ =306.2\mathrm{N} \end{gathered}$$

Hence, thex-component of the force on the block due to rope 1 is 306.2N .