Question

As shown in the figure a block of mass \(10\, kg\) lying on a horizontal surface is pulled by a force \(F\)acting at an angle \(30^{\circ}\), with horizontal For \(\mu_{ s }=0.25\), the block will just start to move for the value of \(F\) : [Given g=10ms^-2]

• A. $20\, N$
• B. $35.7\, N$
• C. $33.3 \,N$
• D. $25.2\, N$

Hint:

Resolve the applied force into its components and carefully consider the forces acting on the block in both the horizontal and vertical directions. The condition for impending motion is that the applied force equals the maximum static friction force.

Answer 1

The Correct Option is D

Step 1: Resolve the Force F into Components
The force F can be resolved into horizontal (\( F \cos 30^\circ \)) and vertical (\( F \sin 30^\circ \)) components.

Step 2: Calculate the Normal Force
The normal force (N) acting on the block is given by:
\( N = Mg - F \sin 30^\circ \)
where M is the mass of the block and g is the acceleration due to gravity. Given \( M = 10 \, \text{kg} \) and \( g = 10 \, \text{m/s}^2 \), we have:
\( N = 10 \times 10 - F \times \frac{1}{2} = 100 - \frac{F}{2} \).

Step 3: Apply the Condition for Impending Motion
The block will just start to move when the horizontal component of the applied force equals the maximum static friction force:
\( F \cos 30^\circ = \mu_s N \)
where \( \mu_s \) is the coefficient of static friction. Given \( \mu_s = 0.25 \), we have:
\( F \frac{\sqrt{3}}{2} = 0.25 \left( 100 - \frac{F}{2} \right) \)
\( \frac{\sqrt{3} F}{2} = 25 - \frac{F}{8} \)
Multiplying both sides by 8, we get:
\( 4 \sqrt{3} F = 200 - F \)
\( F(4 \sqrt{3} + 1) = 200 \)
\( F = \frac{200}{4 \sqrt{3} + 1} \approx \frac{200}{4(1.732) + 1} \approx \frac{200}{7.928} \approx 25.22 \, \text{N} \).

Conclusion: The block will just start to move when \( F \) is approximately \( 25.2 \, \text{N} \) (Option 4).