RBD CYU 13 - CHATUP707

Solution 13 – Rolling Sphere

Solution for Question 13

Step 1: Force Analysis

We apply a force $F$ at an angle $\theta$ with the vertical to push the sphere against the wall (providing Normal force) and roll it upwards (overcoming Gravity and Friction).

Horizontal Forces: The horizontal component of $F$ ($F \sin \theta$) pushes against the wall and is balanced by the normal reaction $N$. $$ N = F \sin \theta $$ Vertical Forces: The vertical component of $F$ ($F \cos \theta$) acts upwards. It must overcome the weight ($mg$) and the limiting friction ($f = \mu N$) which acts downwards. $$ F \cos \theta = mg + \mu N $$

Step 2: Formulate the Expression for F

Substitute $N = F \sin \theta$ into the vertical equilibrium equation: $$ F \cos \theta = mg + \mu (F \sin \theta) $$ Rearrange to isolate $F$: $$ F (\cos \theta – \mu \sin \theta) = mg $$ $$ F = \frac{mg}{\cos \theta – \mu \sin \theta} $$

Step 3: Minimize F

To find the minimum force $F_{min}$, we must maximize the denominator $D(\theta) = \cos \theta – \mu \sin \theta$. Using the auxiliary angle method, $A \cos \theta – B \sin \theta \le \sqrt{A^2 + B^2}$. The maximum value of the denominator is $\sqrt{1 + \mu^2}$.

The minimum force required is: $$ F_{\text{min}} = \frac{mg}{\sqrt{1 + \mu^2}} $$