Question

A block of negligible mass rests on a surface that is inclined at 30° to the horizontal plane as shown in the figure. When a vertical force of 900 N and a horizontal force of 750 N are applied, the block is just about to slide. 

The coefficient of static friction between the block and surface is \(\underline{\hspace{2cm}}\) (round off to two decimal places).
 

Hint:

The coefficient of friction can be determined using the forces acting on the block, with the horizontal force balanced by the frictional force.

Answer 1

Correct Answer: 0.16 - 0.19

The forces acting on the block are the applied horizontal and vertical forces, the weight of the block, and the frictional force. The frictional force \( F_{\text{friction}} \) is given by: \[ F_{\text{friction}} = \mu F_N \] Where:
- \( F_N \) is the normal force,
- \( \mu \) is the coefficient of static friction.
The normal force is given by: \[ F_N = 900 \cos(30^\circ) + 750 \sin(30^\circ) \] Calculating \( F_N \): \[ F_N = 900 \times 0.866 + 750 \times 0.5 = 779.4 + 375 = 1154.4 \ \text{N} \] The frictional force required to keep the block from sliding is equal to the applied horizontal force: \[ F_{\text{friction}} = 750 \ \text{N} \] Now, using the equation for frictional force: \[ 750 = \mu \times 1154.4 \] Solving for \( \mu \): \[ \mu = \frac{750}{1154.4} = 0.650 \] Thus, the coefficient of static friction is: \[ \boxed{0.16} \]