Fm Magnitude in 90° Arm Holding 60-N Weight

[arram]

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

 

[Tide]

HINT: The sum of the torques about the pivot point must add to zero.

 

[quicknote]

I would approach this problem by first identifying the known variables and using them to calculate the unknown variable. From the given information, we know that there is a 60-N weight being held in the hand and that the forearm is making a 90° angle with the upper arm. We also know that the biceps muscle is exerting a force, Fm, at a distance of 3.4 cm from the pivot point at the elbow joint. The distance from the weight to the pivot point is also given as 30 cm.

To solve for the magnitude of Fm, we can use the equation for torque: τ = Fd, where τ is torque, F is force, and d is the distance from the pivot point. In this case, the torque is equal to the weight of the object (60 N) multiplied by the distance from the pivot point (30 cm). Therefore, we can set up the equation: τ = (60 N)(30 cm). We also know that the torque is equal to the force exerted by the biceps muscle (Fm) multiplied by its distance from the pivot point (3.4 cm). So we can set up another equation: τ = Fm(3.4 cm).

Since both equations are equal to the torque, we can set them equal to each other and solve for Fm: (60 N)(30 cm) = Fm(3.4 cm). Solving for Fm, we get a magnitude of approximately 510 N for the force exerted by the biceps muscle.

To further visualize this problem, we can draw a free body diagram as shown in the figure provided. The weight of the object is represented by a downward arrow with a magnitude of 60 N. The force exerted by the biceps muscle, Fm, is represented by an upward arrow with a magnitude of 510 N. The pivot point, O, is shown as a dot and the distances of 3.4 cm and 30 cm are labeled on the diagram. This free body diagram helps us understand the forces acting on the arm and how they are balanced to keep the arm in equilibrium.

In conclusion, the magnitude of Fm in this scenario is approximately 510 N. It is important to note that this is a simplified calculation and does not take into account factors such as the weight of the arm and hand, as well as the tension in other muscles that may also be contributing