Question:
Consider a block of mass \(m\) hanging using a string and pulley arrangement, as shown in the figure. The weight \(mg\) and tension \(T\) are working on the block in such a way that the block is not moving and the string is parallel to the perfectly vertical wall. If the block is just in contact but not attached/fixed with the wall and the coefficient of static friction is \(\mu\), then the static frictional force acting on the block is
Consider a block of mass \(m\) hanging using a string and pulley arrangement, as shown in the figure. The weight \(mg\) and tension \(T\) are working on the block in such a way that the block is not moving and the string is parallel to the perfectly vertical wall. If the block is just in contact but not attached/fixed with the wall and the coefficient of static friction is \(\mu\), then the static frictional force acting on the block is
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Friction depends on normal reaction. If the normal reaction is zero, then frictional force is also zero, even if the surfaces are in contact.
Updated On: Jun 5, 2026
- \(0\)
- \(\mu mg\)
- \(\mu T\)
- \(\mu \left(\dfrac{mg+T}{2}\right)\)
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The Correct Option is A
Solution and Explanation
Step 1: Identify the forces on the block.
The block is acted upon by weight \(mg\) vertically downward and tension \(T\) vertically upward.
Step 2: Observe the direction of the string.
The string is parallel to the vertical wall, so tension has no horizontal component.
Step 3: Check horizontal forces.
Since there is no horizontal pull or push on the block, the wall does not exert any normal reaction on the block.
Step 4: Write the normal reaction.
\[ N=0 \]
Step 5: Use formula for limiting friction.
Static friction can exist only when there is normal reaction. Its maximum value is
\[ f_s \leq \mu N \]
Step 6: Substitute \(N=0\).
\[ f_s \leq \mu \cdot 0 \] \[ f_s=0 \]
Step 7: Final conclusion.
Since the block is only just in contact with the wall and no normal force acts on it, the static frictional force is zero.
\[ \boxed{0} \]
Hence, the correct answer is option (A).
The block is acted upon by weight \(mg\) vertically downward and tension \(T\) vertically upward.
Step 2: Observe the direction of the string.
The string is parallel to the vertical wall, so tension has no horizontal component.
Step 3: Check horizontal forces.
Since there is no horizontal pull or push on the block, the wall does not exert any normal reaction on the block.
Step 4: Write the normal reaction.
\[ N=0 \]
Step 5: Use formula for limiting friction.
Static friction can exist only when there is normal reaction. Its maximum value is
\[ f_s \leq \mu N \]
Step 6: Substitute \(N=0\).
\[ f_s \leq \mu \cdot 0 \] \[ f_s=0 \]
Step 7: Final conclusion.
Since the block is only just in contact with the wall and no normal force acts on it, the static frictional force is zero.
\[ \boxed{0} \]
Hence, the correct answer is option (A).
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