A skier is currently being pulled up the mountain by a snowmobile at a constant

Question

A skier is currently being pulled up the mountain by a snowmobile at a constant velocity. The rope has a tension of 70 N, with direction, e_T= 0.236i + 0.943j + 0.236k. The skis are pointed in a direction other than the rope, given by e_p = 0i + 0.992j + 0.124k and the normal vector is e_n = 0i - 0.124j + 0.992k. The position vector from the person's hands to his right boot is r_AB = 0.25i + 0.5j - 1,0k meters. Assume there is no friction in the direction parallel to the skis. Find the weight of the skier. Now assume the normal force on the skier's left boot is zero, hind the.v component of the position vector, R_B,cg = [x, y, z], from the right boot to the CG of the skier.

Explanation / Answer

eT = (0.236i +0.943j +0.236k)

T    =(0.236i +0.943j +0.236k) *70 N

      = 16.52i + 66.01 j + 16.52 k

ep = 0i +0.992j + 0.124k

en   = 0i -0.124j +0.992 k

weight of the skier acts vertically downward

Fw = 0i+0j-wk

       = wk

Now the only forces acting on the skier are

tension in the rope given by eT and his weight ew and normal reaction N of the mountain plane along en

normal reaction Fn = 0i -0.124Nj +0.992N k

As the skier is moving at constant velocity along ep algebraic some of all the forces must be 0 along ep

T = 16.52i + 66.01 j + 16.52 k

component of T along ep

ep .T = (0i +0.992j + 0.124k) . (16.52i + 66.01 j + 16.52 k)

= 66.01*0.992 +16.52*0.124 = 67.53

Tp = (0i +0.992j + 0.124k)*67.53

     = 66.99j + 8.37 k

    

Fw = -wk

Fn = -0.124Nj +0.992N k

Tp +Fw +Fn =0

0.992N-w+8.37 =0

0.124N = 66.99

N = 66.99/0.124 = 540.24

w = 0.992*540.24+8.37 = 544.29

mass of the skier = 544.29/9.8 = 55.54 kg

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