The coefficient of static friction between a 40kg crate and concrete surface is 0.25. Find the magnitude of the minimum force needed to keep the crate stationary on the concrete base inclined at 45o to the horizontal. [G = 10ms-1]
- 400N
- 300N
- 283N
-
212N
The correct answer is: D
Explanation
momentum force = Rsin\(\theta\) - \(\mu\)Rcos\(\theta\)
Fm = (40 x 10 x sin 45) - (40 x 10 x 0.25 x cos 45)
= 212N