A 100 kg device is pulled up a plane inclined at 30^\circ 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]
A
283.5N
B
975.0N
C
216.5N
D
716.5N
Verified Correct Answer: D.
Physical Principles & Numerical Solution
The forces opposing the motion of the device are "mgsin \theta" and "\muN" where N is the normal reaction
From the diagram, on resolving to components, N = mgcos \theta
Total opposing force = mgsin \theta + \mumgcos \theta = mg(sin \theta + \mucos \theta)
= 100 \times 10 \times (sin 30^\circ + 0.25 \times cos 30^\circ)
= 1000 \times (0.5 + 0.2165)
= 1000 \times 0.7165
= 716.5 N
Source: Myschool Past Questions Archive (Record #69369)