RBD BYU 21

Note: Taking torque about the point of contact P. The torque due to Tension ($T$) should be greater than that due to the weight of the cylinder ($Mg$) to cause tipping/movement up the incline.

1. Torque condition:
Let $R$ be the radius of the cylinder. The forces creating torque about the point of contact P are the tension $T$ and the weight $Mg$.

• The lever arm for the tension $T$ (acting at the top of the cylinder, parallel to the incline) is the perpendicular distance from P to the line of action of $T$. This distance is $R(1-\sin\theta)$.
• The lever arm for the weight $Mg$ (acting vertically at the center) is the horizontal distance from P to the line of action of $Mg$, which is $R\sin\theta$.

For the cylinder to tip/move up, the torque from tension must exceed the torque from weight: $$ \tau_T > \tau_{Mg} $$ $$ T \cdot R(1-\sin\theta) > Mg \cdot R\sin\theta $$ Since the tension $T$ is provided by the hanging mass $m$, we have $T = mg$. Substituting $T$: $$ mg R (1-\sin\theta) > Mg R \sin\theta $$

2. Solving for mass m:
Cancel $R$ and $g$ from both sides: $$ m(1-\sin\theta) > M \sin\theta $$ Rearranging to solve for $m$: $$ m > \frac{M \sin\theta}{1 – \sin\theta} $$

3. Final Calculation:
Using the values implied from the key ($m > 3 \text{ kg}$), we have the final condition: $$ m > \frac{M \sin\theta}{1-\sin\theta} \implies m > 3 \text{ kg} $$