Question

A block of weight 1200N is on an incline plane of 30° with the horizontal, a force P is applied to the body parallel to the plane, if the coefficient of the static friction is 0.20 and kinetic friction is 0.15 (1) find the value of P to cause motion up the plane (2) find P to prevent motion down the plane. (3) Find P to cause continuous motion up the plane.​

Answer 1

a) P = 807.85 N, b) P = 392.15 N, c) P = 444.12 N

Step-by-step explanation:

For this exercise, let's use Newton's second law, let's set a reference frame with the x-axis parallel to the plane and the direction rising as positive, and the y-axis perpendicular to the plane.

Let's use trigonometry to break down the weight

sin θ = Wₓ / W

cos θ = W_y / W

Wₓ = W sin θ

W_y = W cos θ

Wₓ = 1200 sin 30 = 600 N

W_y = 1200 cos 30 = 1039.23 N

Y axis

N- W_y = 0

N = W_y = 1039.23 N

Remember that the friction force always opposes the movement

a) in this case, the system will begin to move upwards, which is why friction is static

P -Wₓ -fr = 0

P = Wₓ + fr

as the system is moving the friction coefficient is dynamic

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = 600+ 207.85

P = 807.85 N

b) to avoid downward movement implies that the system is stopped, therefore the friction coefficient is static

P + fr -Wx = 0

fr = μ N

fr = 0.20 1039.23

fr = 207.85 N

we substitute

P = Wₓ -fr

P = 600 - 207,846

P = 392.15 N

c) as the movement is continuous, the friction coefficient is dynamic

P - Wₓ + fr = 0

P = Wₓ - fr

fr = 0.15 1039.23

fr = 155.88 N

P = 600 - 155.88

P = 444.12 N