Question

At the presentation ceremony, a championship bowler is presented a 1.60-kg trophy which he holds at arm's length, a distance of 0.655 m from his shoulder joint.
(a) Determine the torque the trophy exerts about the shoulder joint when his arm is horizontal. (Enter the magnitude only.)

(b) Determine the torque the trophy exerts about the shoulder joint when his arm is at an angle of 15.0

Answer 1

Answer:

a)10.28 Nm

b)9.93 Nm

Explanation:

Let g = 9.81m/s2. First we can calculate the weight of the trophy

W = mg = 1.6 * 9.81 = 15.696 N

(a) The torque is product of force and its moment arm

T = WL = 15.696 * 0.655 = 10.28 Nm

(b) Suppose his arm makes an angle of 15 degree with respect to the horizontal line. We can still calculate the arm length, or the horizontal distance from the trophy to the champion:

$L_2 = Lcos(15^0) = 0.655cos(15^0) = 0.655*0.966 = 0.633 m$

Again, torque is product of force and its moment arm

$T_2 = WL_2 = 15.696 * 0.633 = 9.93 Nm$