For equilibrium on the ramp, the sum of the forces parallel to the ramp must be 0. For a ramp inclined 30 degrees, the force component of the box parallel to the ramp is 215N * sin(30).
The frictional force is given as 27N up the ramp. Solve for the additional force up the ramp for equilibrium:
F=needed force for equilibrium
positive is up the ramp
-(215N * sin(30)) + 27N + F = 0
The value for F here is a force parallel to the ramp. If the equilibrium force is applied at an angle 40 degrees to the ramp, then solve for the value which has a component parallel to the ramp equal to the needed force above.
F2=force applied at 40 degrees to ramp
F = F2 * cos(40)
F was found above, solve for F2
A box with a weight of 215 N rests on a ramp. There is a frictional force of 27 N exerted on the box directed up the ramp. If the ramp is inclined at 30 degrees, what force must be applied at an angle of 40 degrees to the ramp to maintain equilibrium?
Textbook Answer: 80.5 N and 105 N
Please explain thoroughly!
1 answer
1 answer