You are doing exercises on a Nautilus machine in a gym to strengthen your deltoi

Question

You are doing exercises on a Nautilus machine in a

gym to strengthen your deltoid (shoulder) muscles. Your

arms are raised vertically and can pivot around the

shoulder joint, and you grasp the cable of the machine in

your hand 64.0 cm from your shoulder joint. The deltoid

muscle is attached to the humerus 16.0 cm from the

shoulder joint and makes a 12.0? angle with that bone.

a) If you have set the tension in the cable of the machine

to 33.0 N on each arm, what is the tension in each deltoid

muscle if you hold your outstretched arms in place?

b) What then is the magnitude of the force on the

humerus bone at the shoulder joint if you are holding

your are holding your arm in place? (Assume the mass of the arm is 8.0 kg.)

Explanation / Answer

A)

The torques applied from the cable and from the muscle must equal each other, to hold steady.

Angle of cable from horizontal = (125 - 90) = 35 deg.

Horizontal force component applied from cable
= (cos 35) x 33
= 27.03N.

This is applied at 0.64m. from the shoulder joint (the fulcrum or centre of rotation).

Torque = (0.64 x 27.03)
= 17.30N/m.

To hold steady, the muscle must supply equal torque.

The horizontal component of the torque 0.16m. from the fulcrum is the 17.30N/m.,
so the horizontal force component needed
= (17.30 / 0.16)
= 108.128 N.

The muscle force is applied at an angle of 12 degrees from vertical.

Muscle tension required

= 108.128 / (sin 12)

= 520.0672 N ~ 520 N.

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