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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

Respuesta :

Answer:

[tex]\vec \tau = 31.9 N[/tex]

Explanation:

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

[tex]\vec \tau = \vec r \times \vec F[/tex]

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

[tex]r = Lcos\theta [/tex]

[tex]r = 0.56 cos15 = 0.54 m[/tex]

now we have

[tex]\vec \tau = (0.54)(59)[/tex]

[tex]\vec \tau = 31.9 Nm[/tex]

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