Question

Three crates with various contents are being pulled by a 3575-N force across a horizontal, frictionless roller-conveyor system. The group of boxes is accelerating on the conveyor at 1.578 m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Assume that the ropes and force meters are massless.

Answer 1

Concepts and reason

The main concept used to solve this problem is Newton’s second law of motion.

Initially, draw a free-body diagram of the system of masses and force acting on it. Then, use, Newton’s second law of motion to determine the total mass of the system.

Then, draw the free-body diagram of first block and use the Newton’s second law to find the mass of the first block.

Again, draw the free-body diagram of second block and use the Newton’s second law to find the mass of the second block.

Finally, draw the free-body diagram of third block and use the Newton’s second law to find the mass of the third block.

Fundamentals

Newton’s second law of motion states that acceleration of an object is equal to the ratio of net force acting on the object and mass of the object. The net force of the system is zero if there are no external forces acting on the system. The expression of net force acting on an object is,

$F = ma$

Here, m is the mass and a is the acceleration.

a)

The figure 1 shows a free-body diagram of a system of three masses moving on a conveyor system due to the force F acting on them in the rightward direction. the total mass of the system is the sum of all the masses.

The total mass of the system can be given as follows:

$M = {m_1} + {m_2} + {m_3}$

Here, ${m_1}$ , ${m_2}$ , and ${m_3}$ are the masses of crates 1, 2, and 3, respectively.

The expression of net force acting on the system is,

$F = Ma$

Here, M is the total mass of the system and a is the acceleration of the system.

Rearrange the above expression for M.

$M = \frac{F}{a}$

Substitute 3575 N for F and $1.578{\rm{ m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\M = \frac{{3575{\rm{ N}}}}{{1.578{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 2265.525982{\rm{ kg}}\\\\ = 2265.5{\rm{ kg}}\\\end{array}$

b)

The figure 2 shows a free-body diagram of crate 1 moving on a conveyor system due to the tension ${T_1}$ acting on it in the rightward direction.

The expression of net force acting on the system is,

${T_1} = {m_1}a$

Here, ${m_1}$ is the mass of the crate 1 and a is the acceleration of the crate 1.

Rearrange the above expression for ${m_1}$ .

${m_1} = \frac{{{T_1}}}{a}$

Substitute 1381 N for ${T_1}$ and $1.578{\rm{ m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\{m_1} = \frac{{{\rm{1381 N}}}}{{1.578{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 875.158{\rm{ kg}}\\\\ = 875.2{\rm{ kg}}\\\end{array}$

c)

The figure 3 shows a free-body diagram of crate 2 moving on a conveyor system due to the tension ${T_2}$ acting on it in the rightward direction. The tension ${T_1}$ is acting in the left direction.

Refer figure 3, the net force acting on the crate two is as follows:

${F_{{\rm{net}}}} = {T_2} - {T_1}$

The expression of net force acting on the system by Newton’s second law is,

${F_{{\rm{net}}}} = {m_2}a$

Here, ${m_2}$ is the mass of the crate 2 and a is the acceleration of the crate 2.

Substitute ${T_2} - {T_1}$ for ${F_{{\rm{net}}}}$ and rearrange the above expression for ${m_2}$ .

${m_2} = \frac{{{T_2} - {T_1}}}{a}$

Substitute 1381 N for ${T_1}$ , 2273 N for ${T_2}$ , and $1.578{\rm{ m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\{m_2} = \frac{{{\rm{2273 N}} - {\rm{1381 N}}}}{{1.578{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 565.2724{\rm{ kg}}\\\\ = 565.3{\rm{ kg}}\\\end{array}$

d)

The figure 4 shows a free-body diagram of crate 3 moving on a conveyor system due to the force F acting in the rightward direction. The tension ${T_2}$ is acting on it in the leftward direction.

Refer figure 4, the net force acting on the crate three is as follows:

${F_{{\rm{net}}}} = F - {T_2}$

The expression of net force acting on the system by Newton’s second law is,

${F_{{\rm{net}}}} = {m_3}a$

Here, ${m_3}$ is the mass of the crate 3 and a is the acceleration of the crate 3.

Substitute ${T_2} - {T_1}$ for ${F_{{\rm{net}}}}$ and rearrange the above expression for ${m_3}$ .

${m_3} = \frac{{F - {T_2}}}{a}$

Substitute 2273 N for ${T_2}$ , 3575 N for F and $1.578{\rm{ m/}}{{\rm{s}}^2}$ for a in the above expression.

$\begin{array}{c}\\{m_3} = \frac{{{\rm{3375 N}} - {\rm{2273 N}}}}{{1.578{\rm{ m/}}{{\rm{s}}^2}}}\\\\ = 825.095{\rm{ kg}}\\\\ = 825.1{\rm{ kg}}\\\end{array}$

Ans: Part a

The total mass of the three boxes is 2265.5 kg.

Part b

The mass of first crate is 875.2 kg.

Part c

The mass of second crate is 565.3 kg.

Part d

The mass of second crate is 825.1 kg.