Question

Part B

The horizontal force $F_h_vec$ would rotate the femurheadclockwise in the hip socket, but the frictional force $f_f_vec$ acts to prevent thisclockwise rotation. The apparatus is designed such that when, the femur head rotates clockwise. A sample of datacollected at impending slippage of the femur isshown in thefigure.(PartB 1figure)

Based on these data, what is the approximate coefficient of staticfriction $mu_s$ between the femur head and the hipsocket?Enter your answer numerically to onesignificant figure.

Answer 1

Part B)

The normal vector is vertically upward Because the Femur head rotates in clockwise and friction force always opposes the motion. Therefore Friction force is towards forward This is shown below. N^vector = F_d^vector Friction force f_f^vector = mu n^vector = mu F_d^vector if we take Moment at centre point F_n^vector x = F_f^vector x therefore f_f^vector = F_n^vector \r\n F_n^vector = mu F_d^vector mu = F_n^vector/F_d^vector mu_1 = 1.5/175 = 0.00857 mu_2 = 3.75/325 = 0.0115 mu_3 = 5/500 = 0.01 mu_4 = 7/800 = 0.00875 mu_5 = 9/1000 = 0.009

Coefficient of friction is given by the slope of the graph given , hence

$\mu_{s}$ = Slope = Y-intercept/X-intercept

$\mu_{s}$ = $\Delta$Y/$\Delta$X

$\mu_{s}$ = 9/1000

$\mu_{s}$ = 9 x 10^-3

Answer 2

The normal vector is vertically upward Because the Femur head rotates in clockwise and friction force always opposes the motion. Therefore Friction force is towards forward This is shown below. N^vector = F_d^vector Friction force f_f^vector = mu n^vector = mu F_d^vector if we take Moment at centre point F_n^vector x = F_f^vector x therefore f_f^vector = F_n^vector \r\n F_n^vector = mu F_d^vector mu = F_n^vector/F_d^vector mu_1 = 1.5/175 = 0.00857 mu_2 = 3.75/325 = 0.0115 mu_3 = 5/500 = 0.01 mu_4 = 7/800 = 0.00875 mu_5 = 9/1000 = 0.009