Question

A man is holding a 6.0-kg (weight = 59 N) dumbbell at arm's length, a distance of 0.56 mfrom his shoulder. What is the torque on the shoulder joint from the weight of the dumbbell if thearm is held at 15° above the horizontal? On the picture, draw the lever arm for this force

Answer 1

`\vec \tau = 31.9 N`

Step-by-step explanation:

As we know that torque due to a force is given by the formula

`\vec \tau = \vec r * \vec F`

here we know that force is exerted due to weight of the mass hold in his hand

so we have

F = mg = 59 N

Now Lever arm is the perpendicular distance on the line of action of force from the axis about which the system is rotated

so here we can say

`r = Lcos\theta`

`r = 0.56 cos15 = 0.54 m`

now we have

`\vec \tau = (0.54)(59)`

`\vec \tau = 31.9 Nm`