A 100 kg device is pulled up a plane inclined at 30° to the horizontal with a force of 1000 N. If the coefficient of friction between the device and the surface is 0.25, determine the total force opposing the motion.
[ g = 10 ms-2]
- 283.5N
- 975.0N
- 216.5N
-
716.5N
The correct answer is: D
Explanation
The forces opposing the motion of the device are "mgsin θ" and "μN" where N is the normal reaction
From the diagram, on resolving to components, N = mgcos θ
Total opposing force = mgsin θ + μmgcos θ = mg(sin θ + μcos θ)
= 100 × 10 × (sin 30° + 0.25 × cos 30°)
= 1000 × (0.5 + 0.2165)
= 1000 × 0.7165
= 716.5 N