Can a cylinder pull a box

Figure

If the cord is stressed the cylinder and the block are moving with the same acceleration a. Let F be the tension in the cord, S the frictional force between the cylinder and the inclined plane (Fig. 2). The angular acceleration of the cylinder is a/r. The net force causing the acceleration of the block: $m_{2}a=m_{2}g\sin\alpha-\mu m_{2}g\cos\alpha+F$

and the net force causing the acceleration of the cylinder:

$m_{1}a=m_{1}g\sin \alpha -S-F$

The equation of motion for the rotation of the cylinder:

$Sr=\frac {a}{r}.I$ (I is the moment of inertia of the cylinder, S⋅r is the torque of the frictional force.)

Solving the system of equations we get

$a=g\frac{(m_{1}+m_{2})\sin\alpha-\mu m_{2}\cos\alpha }{m_{1}+m_{2}+\frac {I}{r^{2}} }$

$S=\frac {I}{r^{2}}g\frac{(m_{1}+m_{2})\sin\alpha-\mu m_{2}\cos\alpha }{m_{1}+m_{2}+\frac {I}{r^{2}} }$

$F=m_{2}g\frac{(m_{1}+\frac {I}{r^{2}})\sin\alpha-\mu \frac {I}{r^{2}}\cos\alpha }{m_{1}+m_{2}+\frac {I}{r^{2}} }$

The moment of inertia of a solid cylinder is

$I=\frac {m_{1}r^{2}}{2}$

Using the given numerical values

$a=g\frac{(m_{1}+m_{2})\sin\alpha-\mu m_{2}\cos\alpha }{1.5m_{1}+m_{2}}=0.3317g=3.25m/s^{2}$

$S=\frac {m_{1}g}{2}\frac{(m_{1}+m_{2})\sin\alpha-\mu m_{2}\cos\alpha }{1.5m_{1}+m_{2}}=13.01N$

$F=m_{2}g\frac{((1.5\mu \cos \alpha)-0.5\sin\alpha)m_{1} }{1.5m_{1}+m_{2}}=0.192N.$

Discussion (Mainly for those who feel the question is incomplete.)

If the cord is not stretched the bodies move separately. We obtain the limit by inserting F = 0

$\tan\alpha=\mu (1+\frac {m_{1}r^{2}}{I})=3\mu=0.6 》\alpha=30.96\degree$

The condition for the cylinder to slip is that the value of S (calculated from (2) taking the same coefficient of friction) exceeds the value of αµcos1gm. This gives the same value for α3 as we had for α2. The acceleration of the centers of the cylinder and the block is the same:

$g (\sin\alpha-\mu\cos\alpha)$

, the frictional force at the bottom of the cylinder is $\mu m_{1 }g\cos\alpha$ , the peripheral acceleration of the cylinder is

$\mu\frac {m_{1}r^{2}}{I}g\cos\alpha$