Torque with an arm holding a weight

[wallace13]

A man holds a 169 N ball in his hand, with the forearm horizontal (see the drawing). He can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 21.4 N and has a center of gravity as indicated.



(a) Find the magnitude of M.

(b) Find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint.



(Wmuscle*.051)-(21.4*(.089+.051))-(169*.33)= 0

Wmuscle=1152

I got part A correct, but I cannot get part B. I know that The end of the arm (where the ball is) should now be considered the pivot (zero) point.

Here is my attempt:
-(21.4* (.330-.089-.051))+1152- (W elbow*.330)

W elbow= 3503.23N

 

[PhanthomJay]

A man holds a 169 N ball in his hand, with the forearm horizontal (see the drawing). He can support the ball in this position because of the flexor muscle force M, which is applied perpendicular to the forearm. The forearm weighs 21.4 N and has a center of gravity as indicated.



(a) Find the magnitude of M.

(b) Find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint.



(Wmuscle*.051)-(21.4*(.089+.051))-(169*.33)= 0

Wmuscle=1152

I got part A correct, but I cannot get part B. I know that The end of the arm (where the ball is) should now be considered the pivot (zero) point.

Here is my attempt:
-(21.4* (.330-.089-.051))+1152 (you forget to multiply 1152 by the distance to the ball) - (W elbow*.330)

W elbow= 3503.23N

Alternatively, or as a check, you should just sum all vertical forces to be sure they add to zero.

 

[thomate1]

I would first commend the person for their attempt at solving the problem and provide some guidance on how to approach part B of the question.

Firstly, it's important to understand the concept of torque and its relationship to forces and distance. Torque is the measure of a force's tendency to produce rotation around an axis or pivot point. In this case, the pivot point is the elbow joint, and the force being applied is the flexor muscle force M.

To find the magnitude and direction of the force applied by the upper arm bone to the forearm at the elbow joint, we need to use the equation for torque: T = F x d, where T is torque, F is the applied force, and d is the distance from the pivot point.

To solve for the magnitude of the force applied by the upper arm bone, we can rearrange the equation to F = T/d. The torque in this case is equal to the weight of the forearm (21.4 N) multiplied by its distance from the pivot point (0.089 m) plus the weight of the ball (169 N) multiplied by its distance from the pivot point (0.33 m). This gives us a torque of 58.2 Nm.

Plugging this into the equation, we get F = 58.2 Nm / 0.33 m = 176.36 N. This is the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint.

To find the direction of this force, we can use the right-hand rule. If we curl the fingers of our right hand in the direction of the rotation (counter-clockwise in this case), our thumb will point in the direction of the force. In this case, the force will be directed upwards, perpendicular to the forearm.

In conclusion, the magnitude of the force applied by the upper arm bone to the forearm at the elbow joint is 176.36 N, and its direction is upwards. I would also suggest practicing more problems involving torque and forces to further understand the concept and improve problem-solving skills.