The weight of a block on the inclined plane is 500 N and the angle of incline is 30 degrees. What is the magnitude of the force
of friction for static equilibrium? What is the normal force of the surface pushing up on the block?
1 answer:
Answer:
250 N
433 N
Explanation:
N = Normal force by the surface of the inclined plane
W = Weight of the block = 500 N
f = static frictional force acting on the block
Parallel to incline, force equation is given as
f = W Sin30
f = (500) Sin30
f = 250 N
Perpendicular to incline force equation is given
N = W Cos30
N = (500) Cos30
N = 433 N
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If you notice any mistake in my english, please let me know, because i am not native.