Answer:
b) N = 560 N, c) fr = 138.56 N, d) μ = 0.247
Explanation:
a) In the attachment we can see the free body diagram of the system
b) Let's write Newton's second law on the y-axis
N + T_y -W = 0
N = W -T_y
let's use trigonometry for tension
sin θ = T_y / T
cos θ = Tₓ / T
T_y = T sin θ
Tₓ = T cos θ
we substitute
N = W - T sin 30
we calculate
N = 640 - 160 sin 30
N = 560 N
c) as the system goes at constant speed the acceleration is zero
X axis
Tₓ - fr = 0
Tₓ = fr
we substitute and calculate
fr = 160 cos 30
fr = 138.56 N
d) the friction force has the formula
fr = μ N
μ = fr / N
we calculate
μ = 138.56 / 560
μ = 0.247
Answer:
The solution and the explanation are in the Explanation section.
Explanation:
According to the diagram that is in the attached image, the EFFORT force at point A and the load is at O point. The torque due to weight is:
TA = W * (a * cosθ)
The torque due to effort at C point is:
TC = E * (b * cosθ)
The net torque is equal to 0, we have:
Tnet = 0
W * (a * cosθ) - E * (b * cosθ) = 0
From the figure, you can observe that a/b < 1, thus E < W
