Question

What force \(F\) is required to start moving this 10 kg block shown in the figure if it acts at an angle of \(60^\circ\) as shown? (\(\mu_s = 0.6\)).

• A. 22.72 N
• B. 24.97 N
• C. 25.56 N
• D. 27.32 N

Hint:

To find the force required to move an object at an angle, take into account both the normal force and the frictional force, and solve using the static friction formula.

Answer 1

The Correct Option is B

Step 1: Static friction, \( f_s \), is calculated as:

\[ f_s = \mu_s N \]

where \(\mu_s\) is the static friction coefficient and \(N\) is the normal force.

Step 2: The normal force changes due to an applied force \(F\) at \(60^\circ\). The normal force \(N\) is:

\[ N = mg - F \sin 60 \]

Here, \(m = 10 \, \text{kg}\) (block mass) and \(g = 9.8 \, \text{m/s}^2\) (gravity).

Step 3: To move the block, the applied force \(F\) needs to exceed static friction. Therefore:

\[ F = \frac{f_s}{\mu_s} = \frac{10 \times 9.8 \times 0.6}{\cos 60^\circ} = 24.97 \, \text{N}. \]