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

To solve this problem, we need to determine the minimum force \( F \) required to just move the block. This requires us to consider both the horizontal and vertical components of the force, as well as the frictional force resisting the motion.

Step 1: Understanding Forces Acting on the Block

• The block has a mass \( m = 10 \, \text{kg} \), thus the gravitational force is \( mg = 10 \times 10 = 100 \, \text{N} \) acting downward.
• The force \( F \) is applied at an angle of \( 30^\circ \) with the horizontal.
• The coefficient of static friction \( \mu_s = 0.25 \).

Step 2: Resolve the Force into Components

• Horizontal component of the force: \( F_{\text{horizontal}} = F \cos 30^\circ \).
• Vertical component of the force: \( F_{\text{vertical}} = F \sin 30^\circ \).

Step 3: Calculate the Normal Force

The normal force \( N \) is adjusted by the vertical component of the force \( F \):

\(N = mg - F \sin 30^\circ\)

Step 4: Determine the Maximum Static Friction

The maximum static frictional force \( f_{\text{max}} \) is given by:

\(f_{\text{max}} = \mu_s \cdot N\)

Substituting the expression for \( N \):

\(f_{\text{max}} = \mu_s (mg - F \sin 30^\circ)\)

Step 5: Equate to Overcome Friction

For the block to just start moving, the horizontal component of the force should equal the maximum static friction:

\(F \cos 30^\circ = \mu_s (mg - F \sin 30^\circ)\)

Substituting values and solving for \( F \):

• \(F \cdot \frac{\sqrt{3}}{2} = 0.25 (100 - F \cdot \frac{1}{2})\)

Performing the calculations yields:

\(F \approx 25.2 \, \text{N}\)

Thus, the correct answer is that the block will just start to move for the value of \( F \) being \( 25.2 \, \text{N} \).