A skier is currently being pulled up the mountain by a snowm

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.

Solution

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

T    =(0.236i +0.943j +0.236k) *70 N

      = 16.52i + 66.01 j + 16.52 k

e_p = 0i +0.992j + 0.124k

e_n   = 0i -0.124j +0.992 k

weight of the skier acts vertically downward

F_w = 0i+0j-wk

       = wk

Now the only forces acting on the skier are

tension in the rope given by e_T and his weight e_w and normal reaction N of the mountain plane along e_n

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

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

T = 16.52i + 66.01 j + 16.52 k

component of T along e_p

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

= 66.01*0.992 +16.52*0.124 = 67.53

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

     = 66.99j + 8.37 k

    

F_w = -wk

F_n = -0.124Nj +0.992N k

T_p +F_w +F_n =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