A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 24.8° with respect?
A skier is pulled up a slope at a constant velocity by a tow bar. The slope is inclined at 24.8° with respect to the horizontal. The force applied to the skier by the tow bar is parallel to the slope. The skier's mass is 64.3 kg, and the coefficient of kinetic friction between the skis and the snow is 0.113. Calculate the magnitude of the force that the tow bar exerts on the skier.
i got 24
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F = m*g*sin 24.8+m*g*cos 24.8*μ = 64.3*9.8*(0.419+0.908*0.113) = 329 N
F = mg sin(theta) + k mg cos(theta) = mg(sin(theta) + k cos(theta) = 64.3*9.8*(sin(radians(24.8)) + .113*cos(radians(24.8))) = 328.952575 = 329 N. ANS.
Nope, do not concur. As you failed to show your work I've no clue where you went wrong. But take a look at:
The weight and friction terms are they added? The pull force must act against both to pull the skier uphill.
Did you remember to put both the weight and friction force along the surface of the slope? You need both components along the surface; so you need sine and cosine to direct the two forces along the surface.
By F(net) = F(applied) - F(gravity) - F(friction)
As v = constant=>a = 0=>F(net) = 0
=> F(applied) = F(gravity) + F(friction)
=>Fa = Fg + Ff
=>Fa = mgsinθ + µk x N
=>Fa = mgsinθ + µk x mgcosθ
=>Fa = mg(sinθ + µk x cosθ)
=>Fa = 64.3 x 9.8 x (sin24.8* + 0.113 x cos24.8*)
=>Fa = 328.95 Newton