Question

Figure 5-60 shows a box of dirty money (mass ${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{3}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}$) on a frictionless plane inclined at angle ${\mathit{\theta }}_{\mathbf{1}}\mathbf{=}\mathbf{30}\mathbf{°}$. The box is connected via a cord of negligible mass to a box of laundered money (mass ${\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$) on a frictionless plane inclined at angle ${\mathit{\theta }}_{\mathbf{2}}\mathbf{=}\mathbf{60}\mathbf{°}$.The pulley is frictionless and has negligible mass. What is the tension in the cord?

Short answer

Tension in the cord is 16 N

Step-by-step solution

Given information

It is given that,

$$\begin{gathered} m_1=3.0\mathrm{kg} \\ {\theta }_1=30° \\ {m}_2=2.0\mathrm{kg} \\ {\theta }_2=60° \end{gathered}$$

Determining the concept

The problem is based on Newton’s second law of motion which states that the rate of change of momentum of a body is equal in both magnitude and direction of the force acting on it. Also, this problem deals with the resolution of vectors. Since pulley is ideal, the tension across both sides of the string is the same and also the acceleration of both the objects would be same. Thus, using Newton’s second law of motion and concept of vector resolution, the tension can be calculated by writing and solving equation of motion for individual mass.

Formula:

$F_{net}=ma$ (i)

where, $F_{net}$ is the net force, Mis mass and a is an acceleration.

Determining the tension in the cord

Free Body Diagram:

According to Newton’s second law:

$m_{1}g\sin \left({\theta }_1\right)-T=m_{1}a$ (ii)

$T-m_{2}g\sin \left({\theta }_2\right)=m_{2}a$ (iii)

By adding equations(ii) and (iii), T will cancel out to get a in terms of masses.

$m_{1}g\sin \left({\theta }_1\right)-m_{1}g\sin \left({\theta }_1\right)=\left(m_1+m_2\right)a$

Hence,

$$\begin{gathered} a=\frac{m_{1}g\sin \left({\theta }_1\right)-m_{2}g\sin \left({\theta }_2\right)}{\left(m_1+m_2\right)} \\ a=\frac{\left(3\times 9.8\times \sin \left(30\right)\right)-\left(2\times 9.8\times \sin \left(60\right)\right)}{\left(3+2\right)} \\ a=-0.45\mathrm{m}/{\mathrm{s}}^2 \end{gathered}$$

Negative sign indicates that mass $m_1$is moving upward.

Now, by plugging this value in equation (ii),

$$\begin{gathered} 3\times 9.8\times \sin \left(30\right)-T=3\times \left(-0.45\right) \\ 14.7-T=-1.35 \\ T=16\mathrm{N} \end{gathered}$$

Hence, the tension in the cord is$T=16\mathrm{N}$