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

Step 1: Set Up Coordinate System

Let's use a coordinate system aligned with the inclined plane:

• x-axis: along the incline (positive down the slope)
• y-axis: perpendicular to the incline (positive away from surface)

Step 2: Resolve Applied Forces

Vertical force (900 N downward):

• Component perpendicular to incline: $F_{v,\perp} = 900 \cos(30°) = 900 \times 0.866 = 779.42$ N
• Component parallel to incline: $F_{v,\parallel} = 900 \sin(30°) = 900 \times 0.5 = 450$ N (down the slope)

Horizontal force (750 N):

• Component perpendicular to incline: $F_{h,\perp} = 750 \sin(30°) = 750 \times 0.5 = 375$ N (into surface)
• Component parallel to incline: $F_{h,\parallel} = 750 \cos(30°) = 750 \times 0.866 = 649.52$ N (up the slope)

Step 3: Calculate Normal Force

The normal force balances all perpendicular components:

$$N = F_{v,\perp} + F_{h,\perp}$$

$$N = 779.42 + 375 = 1154.42 \text{ N}$$

Step 4: Apply Equilibrium in Parallel Direction

At the point of sliding, the friction force equals $\mu_s N$ and the block is in equilibrium:

For forces parallel to the incline:

• Forces down the slope: $F_{v,\parallel} = 450$ N
• Forces up the slope: $F_{h,\parallel} = 649.52$ N
• Friction force: $f = \mu_s N$ (opposes motion)

Since the horizontal force component up the slope exceeds the vertical force component down the slope, the block tends to move up the incline. Therefore, friction acts down the slope.

$$F_{h,\parallel} = F_{v,\parallel} + f$$

$$649.52 = 450 + \mu_s \times 1154.42$$

$$\mu_s \times 1154.42 = 199.52$$

$$\mu_s = \frac{199.52}{1154.42} = 0.1728$$

Answer: The coefficient of static friction is 0.17 (rounded to two decimal places).