Engineering Mechanics question and answer

1. A block of mass 5 kg rests on a rough inclined plane at 30°. The coefficient of friction is 0.2. Determine whether the block will slide down or remain at rest.

Solution:

Component of weight down the plane:$$W_{\parallel} = mg \sin 30^\circ = 5 \times 9.81 \times 0.5 = 24.525 \text{ N}$$

Normal reaction:$$N = mg \cos 30^\circ = 5 \times 9.81 \times 0.866 \approx 42.43 \text{ N}$$

Maximum friction force

$$f_{\max} = \mu N = 0.2 \times 42.43 \approx 8.49 \text{ N}$$

Compare forces

• Downward force = 24.53 N
• Maximum opposing friction = 8.49 N

Since:$$24.53 > 8.49$$

Conclusion

The friction is not sufficient to hold the block.