A 10 kg block is placed on a ramp. The ramp has a 20 degree angle above the horizontal. What force of friction is needed to keep the block at rest?
Draw a picture, this is simple.
I am really stuck. I drew a picture and everything.
To determine the force of friction needed to keep the block at rest, we need to consider the forces acting on the block.
First, let's break down the forces acting on the block along the ramp:
1. The weight of the block, represented as F_weight, acts vertically downward. We can calculate the weight using the formula F_weight = mass * gravitational acceleration. In this case, F_weight = 10 kg * 9.8 m/s^2.
2. The force of friction, represented as F_friction, acts parallel to the ramp in the opposite direction of motion. This is the force that we need to calculate.
3. The component of the weight acting along the ramp, represented as F_parallel, counteracts the force of friction. We can calculate F_parallel using the formula F_parallel = F_weight * sin(theta), where theta is the angle of the ramp.
Since the block is at rest, the force of friction must be equal in magnitude but opposite in direction to the component of the weight acting along the ramp (F_parallel). Therefore, F_friction = F_parallel.
Let's plug in the numbers:
F_weight = 10 kg * 9.8 m/s^2 = 98 N
F_parallel = F_weight * sin(20 degrees)
Using a calculator, calculate sin(20 degrees) = 0.3420.
F_parallel = 98 N * 0.3420 = 33.636 N
Therefore, the necessary force of friction to keep the block at rest is 33.636 N.