A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.7 and µk = 0.59, respectively. The acceleration of gravity is 9.8 m/s 2 . 31 kg µ 31◦ What is the frictional force acting on the 31 kg mass? Answer in units of N.

To find the frictional force acting on the 31 kg mass, we first need to determine whether the block is moving or at rest. This will be based on the static friction.

### Step 1: Calculate the gravitational force components
The gravitational force \( F_g \) acting on the block is:

\[
F_g = m \cdot g = 31 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 303.8 \, \text{N}
\]

The component of this force parallel to the incline (\( F_{\parallel} \)) is given by:

\[
F_{\parallel} = F_g \cdot \sin(\theta) = 303.8 \cdot \sin(31^\circ)
\]

\[
F_{\parallel} \approx 303.8 \cdot 0.515 \approx 156.45 \, \text{N}
\]

The component of the gravitational force perpendicular to the incline (\( F_{\perp} \)) is:

\[
F_{\perp} = F_g \cdot \cos(\theta) = 303.8 \cdot \cos(31^\circ)
\]

\[
F_{\perp} \approx 303.8 \cdot 0.857 \approx 260.3 \, \text{N}
\]

### Step 2: Maximum static friction force
The maximum static friction force \( f_s^{\text{max}} \) is given by:

\[
f_s^{\text{max}} = \mu_s \cdot F_{\perp} = 0.7 \cdot 260.3 \approx 182.21 \, \text{N}
\]

### Step 3: Compare the forces
The frictional force needs to balance the parallel component of the gravitational force (\( F_{\parallel} \)) to keep the block at rest. Since the maximum static friction force (182.21 N) is greater than \( F_{\parallel} \) (156.45 N), the block does not move. Therefore, the static friction force is equal to the parallel component of gravity:

\[
f_s = F_{\parallel} = 156.45 \, \text{N}
\]

Thus, the frictional force acting on the 31 kg block is **156.45 N**.

Would you like further details or have any questions? 

Here are five related questions to expand on this topic:
1. How does the angle of the incline affect the frictional force?
2. What happens if the angle is increased beyond a certain point?
3. How would the frictional force change if the mass of the block increased?
4. What would the frictional force be if the block started moving?
5. How does kinetic friction differ from static friction in this context?

**Tip:** Remember that static friction adjusts up to its maximum value to prevent motion, while kinetic friction has a constant value once motion starts.

A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are µs = 0.7 and µk = 0.59, respectively. The acceleration of gravity is 9.8 m/s². 31 kg µ 31° What is the frictional force acting on the 31 kg mass? Answer in units of N.

Learn how to calculate the frictional force acting on a 31 kg block resting on a 31° incline. This detailed solution explains static and kinetic friction, gravitational force components, and how to determine if the block remains at rest. Ideal for students in Grades 11-12 studying physics and mechanics.

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