Problem about tension

[BillyCheung]

Dear all :

I have one question about tension that is following :

The human forearm and hand can be modelled as a beam that is pivoted at one end(the elbow) and supported (by a muscle) at a short distance from the pivot. The beam(forearm) is horizontal, the muscle is vertical, the centre of mass of the beam is 15cm from the pivot, and the point of attachment of the muscle is 4 cm from the pivot. The mass of the beam is 2 kg. The pivot is frictionless.

Calculate the magnitude of the tension in the supporting muscle.

In my idea, torque = r x F = 0.15 x (2)(9.8) = 2.94Nm.

Is the tension in the supporting muscle -2.94Nm?

Thank very much. Good bye

Billy

[jamesrc]

No, the tension in the muscle is a force, not a torque. Balancing the torques about the pivot, you will get that T*4cm = mg*15cm; solve for T.

[BillyCheung]

Thank very much. I understand your idea.

On the other hand, the torque of this case should be (0.04m)(2kg)(9.8ms-2) = 0.784 Nm. Do you agree?

And then what is the force exerted on the forearm by the rest of the arm at the elbow(pivot) if the forearm is stationary and in equilibrium?

Thank a lot. Good Bye

Billy

[Jer]

Torque (moments) is all relative to where you set your datum. So for instance, if you set your datum to be the end from which the dimensions are measured from, you'll get a torque of:

F*d(distance to force from datum)

F = the force due to gravity...i.e. 2kg*9.81m/s^2 = 19.62 N
d = 15 cm = .15 m

So therefore the torque would be 19.62*.15 = 2.943 N*m

[Jer]

Oh, and your acceleration of (9.81 - 2) is kind of odd. I'm not sure where you're getting the -2 from, seeing as how acceleration of gravity will always be 9.81 (generally speaking).

[BillyCheung]

Thank very much. The mass of the object is 2 kg. Good Bye

[BillyCheung]

Last question, how do we calculate the magnitude of the horizontal component of the force exerted on the forearm by the rest of the arm when the foream is horizontal? Thank a lot.