Question

Two masses$m_{\mathrm{A}}=2.0\mathrm{kg}$and$m_{\mathrm{B}}=5.0\mathrm{kg}$are on inclines and are connected together by a string as shown in Fig. 4–61. The coefficient of kinetic friction between each mass and its incline is${\mu }_{\mathrm{k}}=0.30$. If$m_{\mathrm{A}}$moves up, and$m_{\mathrm{B}}$moves down, determine their acceleration. [Ignore masses of the (frictionless) pulley and the cord.]

Short answer

Their acceleration is $-2.2\mathrm{m}/{\mathrm{s}}^2$.

Step-by-step solution

Step 1. Given data and free body diagram

Given data:

The coefficient of kinetic friction is ${\mu }_{\mathrm{k}}=0.30$.

The mass of A is $m_{\mathrm{A}}=2.0\mathrm{kg}$.

The mass of A is $m_{\mathrm{B}}=5.0\mathrm{kg}$.

The angle of the left incline plane is ${\theta }_{\mathrm{A}}=51°$.

The angle of the right incline plane is ${\theta }_{\mathrm{B}}=21°$.

Let T be the tension in the cord and a be the acceleration of the system.

Step 2. Calculation of the friction forces

The friction force on any mass acts against the direction of motion and the tension in the cord remains the same for both masses as a single cord runs over the pulley to hold both masses A and B.

The normal force acting on the mass A is $N_{\mathrm{A}}=m_{\mathrm{A}}g\cos {\theta }_{\mathrm{A}}$.

The friction force on mass A is:

$\begin{array}{c}{f}_{\mathrm{A}}={\mu }_{\mathrm{k}}{N}_{\mathrm{A}}\\ ={\mu }_{\mathrm{k}}{m}_{\mathrm{A}}g\cos {\theta }_{\mathrm{A}}\end{array}$

The normal force acting on mass B is $N_{\mathrm{B}}=m_{\mathrm{B}}g\cos {\theta }_{\mathrm{B}}$.

The friction force on mass B is:

$\begin{array}{c}{f}_{\mathrm{B}}={\mu }_{\mathrm{k}}{N}_{\mathrm{B}}\\ ={\mu }_{\mathrm{k}}{m}_{\mathrm{B}}g\cos {\theta }_{\mathrm{B}}\end{array}$

Step 3. Application of Newton’s law 

Now, applying Newton's second law for mass A, you will get:

$\begin{array}{rcl}T-m_{\mathrm{A}}g\sin {\theta }_{\mathrm{A}}-f_{\mathrm{A}}& =& m_{\mathrm{A}}a\\ T-m_{\mathrm{A}}g\sin {\theta }_{\mathrm{A}}-{\mu }_{\mathrm{k}}{m}_{\mathrm{A}}g\cos {\theta }_{\mathrm{A}}& =& m_{\mathrm{A}}a\dots \left(i\right)\end{array}$

Again, applying Newton's second law for mass B, you will get:

$\begin{array}{rcl}{m}_{\mathrm{B}}g\sin {\theta }_{\mathrm{B}}-f_{\mathrm{B}}-T& =& m_{\mathrm{B}}a\\ m_{\mathrm{B}}g\sin {\theta }_{\mathrm{B}}-{\mu }_{\mathrm{k}}{m}_{\mathrm{B}}g\cos {\theta }_{\mathrm{B}}-T& =& m_{\mathrm{B}}a\dots \left(ii\right)\end{array}$

Step 4. Calculation of acceleration

Now, adding equations (i) and (ii), you will get:

$\begin{array}{c}T-m_{\mathrm{A}}g\sin {\theta }_{\mathrm{A}}-{\mu }_{\mathrm{k}}{m}_{\mathrm{A}}g\cos {\theta }_{\mathrm{A}}+m_{\mathrm{B}}g\sin {\theta }_{\mathrm{B}}-{\mu }_{\mathrm{k}}{m}_{\mathrm{B}}g\cos {\theta }_{\mathrm{B}}-T=m_{\mathrm{A}}a+m_{\mathrm{B}}a\\ m_{\mathrm{B}}\left(\sin {\theta }_{\mathrm{B}}-{\mu }_{\mathrm{k}}\cos {\theta }_{\mathrm{B}}\right)g-m_{\mathrm{A}}\left(\sin {\theta }_{\mathrm{A}}+{\mu }_{\mathrm{k}}\cos {\theta }_{\mathrm{A}}\right)g=a\left(m_{\mathrm{A}}+m_{\mathrm{B}}\right)\end{array}$

After simplification, you will get:

$\begin{array}{c}a=\left[\frac{m_{\mathrm{B}}\left(\sin {\theta }_{\mathrm{B}}-{\mu }_{\mathrm{k}}\cos {\theta }_{\mathrm{B}}\right)-m_{\mathrm{A}}\left(\sin {\theta }_{\mathrm{A}}+{\mu }_{\mathrm{k}}\cos {\theta }_{\mathrm{A}}\right)}{m_{\mathrm{A}}+m_{\mathrm{B}}}\right]g\\ =\frac{\left(5.0\mathrm{kg}\right)\left(\sin 21°-0.30\cos 21°\right)-\left(2.0\mathrm{kg}\right)\left(\sin 51°+0.30\cos 51°\right)}{5.0\mathrm{kg}+2.0\mathrm{kg}}\times 9.80\mathrm{m}/{\mathrm{s}}^2\\ =\frac{0.39-1.93}{7.0}\times 9.80\mathrm{m}/{\mathrm{s}}^2\\ =-2.17\mathrm{m}/{\mathrm{s}}^2\approx -2.2\mathrm{m}/{\mathrm{s}}^2\end{array}$

Hence, their acceleration is $-2.2\mathrm{m}/{\mathrm{s}}^2$.