Coefficient of Static Friction on a 37° Inclined Plane
Question: A block is kept on a rough inclined plane (angle \(37^\circ\)). It remains at rest when a force of \(1\,\text{N}\) acting down the plane is applied. However, when a force is applied up the plane, the block just starts to move when the force reaches \(11\,\text{N}\). (Both forces act parallel to the plane.) Determine the coefficient of static friction \(\mu\) between the block and the plane.
Step‑by‑Step Explanation
1. Determine Weight Components:
For a block of weight \(mg\) on an inclined plane at \(37^\circ\):
- Component down the plane: \(mg\sin 37^\circ\)
- Normal force: \(N = mg\cos 37^\circ\)
2. Friction Force:
Maximum static friction is \(f_{\text{max}} = \mu mg\cos 37^\circ\).
3. Two Limiting Cases (Equilibrium):
(a) When pushing down the plane:
The gravitational component plus an extra \(1\,\text{N}\) is balanced by friction:
\[
mg\sin 37^\circ + 1 = \mu mg\cos 37^\circ.
\]
(b) When pushing up the plane:
The friction plus the gravitational component (acting down) balances an applied \(11\,\text{N}\) force:
\[
11 - mg\sin 37^\circ = \mu mg\cos 37^\circ.
\]
4. Equate the Two Cases:
Setting the right sides equal:
\[
mg\sin 37^\circ + 1 = 11 - mg\sin 37^\circ.
\]
Solve this:
\[
2\,mg\sin 37^\circ = 10 \quad \Longrightarrow \quad mg\sin 37^\circ = 5\,\text{N}.
\]
5. Find \(\mu\):
Substitute back into one of the equilibrium equations:
\[
mg\sin 37^\circ + 1 = \mu mg\cos 37^\circ.
\]
Since \(mg\sin 37^\circ = 5\,\text{N}\),
\[
5 + 1 = \mu mg\cos 37^\circ \quad \Longrightarrow \quad \mu mg\cos 37^\circ = 6\,\text{N}.
\]
Using standard values \(\sin 37^\circ \approx 0.6\) and \(\cos 37^\circ \approx 0.8\), the exact value of \(mg\) cancels in the comparison. The result is
\[
\mu \approx \frac{6}{mg\cos 37^\circ} \approx 0.9.
\]
Final Answer: \(\mu \approx 0.9\).
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