A block weighing 10.0 newtons is on a ramp inclined at 30.0° to the horizontal. A
3.0-newton force of friction, Ff, acts on the block as it is pulled up the ramp at constant velocity with force F, which is parallel to the ramp. What is the magnitude of force F?
The answer given is 8 N. Please explain why this is the answer?
3.0-newton force of friction, Ff, acts on the block as it is pulled up the ramp at constant velocity with force F, which is parallel to the ramp. What is the magnitude of force F?
The answer given is 8 N. Please explain why this is the answer?
-
Because the block has constant velocity, a=0 and the net forces parallel to the ramp must total zero.
F(grav,parallel) + F(f) + F = 0
F = -F(grav,parallel) - F(f)
The component of the gravitational force that pulls the block down the ramp is equal to the gravitational force multiplied by sin(theta). I will take "down the ramp" to be negative and "up the ramp" to be positive.
F(grav,parallel) = F(g) sin(theta)
F(grav,parallel) = -10N * 0.5
F(grav,parallel) = -5N
Since the block is moving up the ramp, friction must act opposite to the motion (down the ramp)
F = 5N + 3N
F = 8N
F(grav,parallel) + F(f) + F = 0
F = -F(grav,parallel) - F(f)
The component of the gravitational force that pulls the block down the ramp is equal to the gravitational force multiplied by sin(theta). I will take "down the ramp" to be negative and "up the ramp" to be positive.
F(grav,parallel) = F(g) sin(theta)
F(grav,parallel) = -10N * 0.5
F(grav,parallel) = -5N
Since the block is moving up the ramp, friction must act opposite to the motion (down the ramp)
F = 5N + 3N
F = 8N