Physics Friction question, please help! 10 points best answer?

Question

Please show all work. Thank you.

A 60N toolbox is dragged horizontally at constant speed by a rope making an angle of 35 degrees with the floor. The tension in the rope is 40N. Determine the magnitude of the friction force and the normal force.

Answer 1

Hello Megan : You have:

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W = 60 N



N = W = 60 N <-------[ acts upward ]



P = 40 N



B = 35 degrees



P sub x = ( P ) ( cos B )



P sub x = ( 40 ) ( cos 35 ) = 32.77 N



P sub x = F sub F = 32.77 N <-------

Answer 2

a) If you drew an appropriate free body diagram and focused only on the horizontal forces acting on the toolbox:



F(net) = Tx - F(friction), F(net) = ma and Tx = Tcos(theta)



Given that the toolbox is moving at a constant speed, the acceleration a is 0; therefore,



Tcos(theta) = F(friction); given that the tension T is 40 N and the angle theta is 35 degrees,



F(friction) = 40cos(35 degrees) = 32.77 N





b) Now note the vertical forces acting on the toolbox:



mg = Ty + F(normal), Ty = Tsin(theta); given that the weight mg is 60 N,



60 = 40sin(35 degrees) + F(normal)



F(normal) = 37.06 N

Answer 3

In vector notation you have f = MA = 0 = P - F; where P is the pull force and F is the friction force. So we convert these into XY components and solve.



Px = P cos(theta) = 40*cos(radians(35)) = ? Newtons; Fx = kN = k (W - Py) where W = the weight = 60 N.

Py = 40*sin(radians(35)) = ? Newtons; Fy = 0. N = (W - Py) is the normal force (weight reduced by upward pull).



So we see that Px = 40*cos(radians(35)) = k (60 - 40*sin(radians(35))) = Fx and then



k = 40*cos(radians(35))/(60 - 40*sin(radians(35))) = 0.884208991 ANS.

And the normal is 60 - 40*sin(radians(35)) = 37.05694255 N. ANS.