Fm Magnitude in 90° Arm Holding 60-N Weight

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The discussion focuses on calculating the force exerted by the biceps muscle (Fm) when holding a 60-N weight at a 90° angle with the forearm. The distance from the weight to the elbow joint pivot point is 30 cm, while the biceps muscle exerts force 3.4 cm from the pivot. To find Fm, the principle of torque equilibrium is applied, stating that the sum of the torques around the pivot must equal zero. The calculated magnitude of Fm is determined using the formula τ = r × F, where τ represents torque, r is the distance from the pivot, and F is the force.

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arram
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A 60-N weight is held in the hand with the forearm making a 90° angle with the upper arm as shown in the figure below (note: the arm is held steady and is not moving). The biceps muscle exerts a force Fm that is 3.4 cm from the pivot point "O" at the elbow joint. Neglecting the weight of the arm and hand, what is the magnitude of Fm if the distance from the weight to the pivot point is 30 cm? Draw a free body diagram. http://img442.imageshack.us/img442/4935/free0in.jpg

can anyone help me on this one? thanks
 
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HINT: The sum of the torques about the pivot point must add to zero.
 

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