Question

In Fig. 6-59, block 1 of mass${\mathit{m}}_{\mathbf{1}}\mathbf{=}\mathbf{2}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}$and block 2 of mass${\mathit{m}}_{\mathit{2}}\mathit{=}\mathbf{1}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{kg}\mathbf{}$are connected by a string of negligible mass. Block 2 is pushed by force of magnitude 20 Nand angle$\mathit{\theta }\mathbf{=}\mathbf{35}\mathbf{°}$. The coefficient of kinetic friction between each block and the horizontal surface is 0.20. What is the tension in the string?

Short answer

Tension in the string is 9.4 N

Step-by-step solution

Given

Mass of block 1,$m_1=2.0\mathrm{kg}$.

Mass of block 2,$m_2=1.0\mathrm{kg}$.

Force applied on block 2,F=20 N .

Angle at which force is applied,$\theta =30.0°$.

Understanding the concept

The problem deals with the Newton’s laws of motion which describe the relations between the forces acting on a body and the motion of the body.Applying Newton's laws, we can solve the given problem.

Calculate the tension in the string

For the$m_2==1.0\mathrm{kg}$block, application of Newton's laws results in

$$\begin{gathered} F\cos \theta -T-f_k=m_{2}axaxis \\ {F}_N-F\sin \theta -m_{2}g=0yaxis \end{gathered}$$

Since $f_k={\mu }_k{F}_N$, these equations can be combined into an equation to solve for:

$F\left(\cos \theta -{\mu }_k\sin \theta \right)-T-{\mu }_k{m}_{2}g=m_{2}a$

Similarly (but without the applied push) we analyze the$m_1=2.0kg$block:

$$\begin{gathered} T-{f^{\text{'}}}_k=m_{1}axaxis \\ {F^{\text{'}}}_N=m_{1}g=0yaxis \end{gathered}$$

Using $f_k={\mu }_k{F^{\text{'}}}_N$, the equations can be combined:

$T-{\mu }_k{m}_{1}g=m_{1}a$

Subtracting the two equations for and solving for the tension, we obtain

$$\begin{gathered} T=\frac{m_1\left(\cos \theta -{\mu }_k\sin \theta \right)}{m_1+m_2}F \\ \Rightarrow T=\frac{\left(2.0\mathrm{kg}\right)\left[\cos 35°-\left(0.20\right)\sin 35°\right]}{2.0\mathrm{kg}+1.0\mathrm{kg}}\left(20\mathrm{N}\right) \\ \Rightarrow T=9.4N \end{gathered}$$