Question

Two blocks with masses m1 and m2 are connected by a massless string over a frictionless pulley. Block 1 sits on a frictionless horizontal surface and block 2 sits on a plane inclined at an angle θ above the horizontal. The coefficient of friction between block 2 and the incline is µk. The pulley, which is a uniform disk, has a mass mp and a radius R. When you release the blocks, both blocks slide without the string slipping on the pulley. In terms of the given parameters and physical constants, what is the magnitude of the acceleration of the blocks?

Answer 1

To find such acceleration, we must first perform a dynamic study of each mass. Performing a free-body diagram for each item (considering that the mass 1 is raised while the mass 2 slides down the inclined platform, and the horizontal surface where is m1 is parallel to the plataform horizontal surface) we have:

1) For m1: The normal force is not consider because the mass 1 is already above the ground level.

   $\sum F_{x}\rightarrow 0$   

$\sum F_{y}\rightarrow T_{1}-W_{1}= m_{1}a$

2) For m2:

$\sum F_{x}\rightarrow m_{2}a+W_{2}\sin \Theta = -F_{f}\rightarrow m_{2}a= W_{2}\sin \Theta-\mu _{k}N_{2}$

$\sum F_{y}\rightarrow N_{2}+T_{1}= W_{2}\cos \Theta$

We have, then, the next expressions:

$\left\{\begin{matrix} T_{1}-W_{1}= m_{1}a\\ N_{2}+T_{1}= W_{2}\cos \Theta\\ m_{2}a= W_{2}\sin \Theta-\mu _{k}N_{2} \end{matrix}\right.$ $\rightarrow \left\{\begin{matrix} T_{1}-m_{1}g= m_{1}a\\ N_{2}+T_{1}= m_{2}g\cos \Theta\\ m_{2}a= m_{2}g\sin \Theta-\mu _{k}N_{2} \end{matrix}\right.$ $\rightarrow \left\{\begin{matrix} T_{1}-m_{1}g= m_{1}a\\ T_{1}= m_{2}g\cos \Theta-N_{2}\\ \frac{m_{2}g\sin \Theta-m_{2}a}{\mu _{k}}=N_{2} \end{matrix}\right.$

$\rightarrow \left\{\begin{matrix} T_{1}-m_{1}g= m_{1}a\\ T_{1}= m_{2}g\cos \Theta-\frac{m_{2}g\sin \Theta-m_{2}a}{\mu _{k}}\\ \end{matrix}\right.$    $\rightarrow \left\{\begin{matrix} m_{2}g\cos \Theta-\frac{m_{2}g\sin \Theta-m_{2}a}{\mu _{k}}-m_{1}g= m_{1}a\\ \\ \end{matrix}\right.$

Now, we can find the aceleration as:

$m_{2}g\cos \Theta-\frac{m_{2}g\sin \Theta-m_{2}a}{\mu _{k}}-m_{1}g= m_{1}a$

$\rightarrow m_{2}g\cos \Theta-m_{1}g= m_{1}a+\frac{m_{2}g\sin \Theta-m_{2}a}{\mu _{k}}$

$\rightarrow \mu _{k}(m_{2}g\cos \Theta-m_{1}g)= \mu _{k}m_{1}a+m_{2}g\sin \Theta-m_{2}a$

$\rightarrow \mu _{k}(m_{2}g\cos \Theta-m_{1}g)-m_{2}g\sin \Theta= a(\mu _{k}m_{1}-m_{2})$

$\mathbf{\rightarrow a= \frac{\mu _{k}(m_{2}g\cos \Theta-m_{1}g)-m_{2}g\sin \Theta}{(\mu _{k}m_{1}-m_{2})} }$