Question

In Fig. 5-61, a tin of antioxidants $\mathbf{(}{\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}\mathbf{)}$on a friction less inclined surface is connected to a tin of corned beef$\mathbf{(}{\mathit{m}}_{\mathbf{2}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathit{k}\mathit{g}\mathbf{)}$.The pulley is mass less and friction less. An upward force of magnitude$\mathit{F}\mathbf{=}\mathbf{6}\mathbf{.}\mathbf{0}\mathbf{}\mathit{N}$acts on the corned beef tin, which has a downward acceleration of$\mathbf{5}\mathbf{.}\mathbf{5}\mathbf{}\mathit{m}\mathbf{/}{\mathit{s}}^{\mathbf{2}}$. What are (a) the tension in the connecting cord and (b) angle $\mathit{\beta }$?

Short answer

a)The tension in the connecting rod is $T=2.6N$

b)Angle is $\beta ={17}^{\circ }$

Step-by-step solution

Given information

Given that,

$m_1=1.0kg$

role="math" localid="1655520757224" $m_2=2.0kg$

role="math" localid="1655520811170" $F=6.0N$

$a=5.5m/s^2$

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.

Formula:

According to the Newton’s second law of motion,

$F_{net}=\sum _{}^{}Ma$

(a) Determining the tension in the connecting rod

Free Body diagram

$T+m_{1}g\sin \left(\beta \right)=m_{1}a$

$$\begin{gathered} m_{2}g-F-T=m_{2}a \\ (2\times 9.8)-6-T=2\times 5.5 \end{gathered}$$

By solving for above equation,

$T=2.6N$

Hence, the tension in the connecting rod is$T=2.6N$

(b) Determining the angle

Now, plugging this value in equation 1,

$$\begin{gathered} 2.6+1\times 9.8\times \sin \left(\beta \right)=1\times 5.5 \\ \beta ={17}^{\circ } \end{gathered}$$

Hence, an angle is$\beta ={17}^{\circ }$