A wheel of radius `R` has an axle of radius `R//5`. A force `F` is applied tangentially to the wheel. To keep the system in a state of "rotational" rest, a force `F'` is applied tangentially to the axle. The value of `F'` is

Options

A. `F`
B. `3F`
C. `5F`
D. `7F`

Explanation

To solve the problem, we need to analyze the forces and torques acting on the wheel and the axle. Here’s a step-by-step solution:

### Step 1: Understand the system
We have a wheel of radius \( R \) and an axle of radius \( \frac{R}{5} \). A force \( F \) is applied tangentially to the wheel, and we need to find the force \( F' \) that is applied tangentially to the axle to keep the system in a state of rotational rest.

### Step 2: Set up the torque equation
To keep the system in rotational rest, the net torque about any point must be zero. We can choose the center of the wheel (point O) as our pivot point. The torque due to the force \( F \) acting on the wheel is given by:

\[ \text{Torque due to } F = F \cdot R \]

The torque due to the force \( F' \) acting on the axle is given by:

\[ \text{Torque due to } F' = F' \cdot \left(\frac{R}{5}\right) \]

### Step 3: Write the torque balance equation
Since the system is in rotational rest, the total torque about point O must be zero:

\[ F \cdot R - F' \cdot \left(\frac{R}{5}\right) = 0 \]

### Step 4: Solve for \( F' \)
Rearranging the equation gives:

\[ F \cdot R = F' \cdot \left(\frac{R}{5}\right) \]

Now, we can cancel \( R \) from both sides (assuming \( R \neq 0 \)):

\[ F = F' \cdot \left(\frac{1}{5}\right) \]

Multiplying both sides by 5, we find:

\[ F' = 5F \]

### Final Answer
Thus, the value of \( F' \) is:

\[ F' = 5F \]

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