Question

At the presentation ceremony, a championship bowler is presented a 1.64-kg trophy which he holds at arm's length, a distance of 0.655 m from his shoulder joint. (a) Determine the torque (in N · m) the trophy exerts about the shoulder joint when his arm is horizontal. (Enter the magnitude only.) N · m (b) Determine the torque (in N · m) the trophy exerts about the shoulder joint when his arm is at an angle of 26.0° below the horizontal. (Enter the magnitude only.)

Answer 1

To solve the problem it is necessary to take into account the concepts of the kinetic equations for the description of the torque at the rate of force and distance.

By definition the torque is given by,

$\tau = F*d$

where,

$F= Force$

$d = Distance$

For the problem in question the mass of the trophy is 1.64Kg and the distance of the tropeo to the board (the shoulder) is 0.655m

PART A) For part A, the torque with the given mass and the stipulated torque in the horizontal plane must be calculated as well,

$\tau = F*d$

For Newton's second law

$\tau = mg*d$

$\tau = 1.64*9.81*0.655$

$\tau = 10.5Nm$

PART B) For part B there is an angle of 26 degrees with respect to the horizontal, therefore to know the net torque it is necessary to know the horizontal component to the formed angle, that is,

$\tau = F*dcos\theta$

$\tau = mgdcos\theta$

$\tau = 1.64*9.81*0.655*cos26$

$\tau = 9.471Nm$