How Much Weight Can a Bodybuilder's Deltoid Muscle Hold Horizontally?

[tascja]

Homework Statement

A bodybuilder holds a dumbbell of weight Wd as shown in Figure 9.8a. His arm is extended horizontally and weighs Wa = 31.0 N. The deltoid muscle is assumed to be the only muscle acting and is attached to the arm as shown. The maximum force M that the deltoid muscle can supply to keep the arm horizontal has a magnitude of 1840 N. Figure 9.8b shows the distances that locate where the various forces act on the arm. What is the heaviest dumbbell that can be held, and what are the horizontal and vertical force components, Sx and Sy that the shoulder joint applies to the arm?

Homework Equations

The Attempt at a Solution

so i already found that the max weight the dumbbell can be is 86.1 N. but I am not sure about the Sx and Sy components. I thought they were the components of the deltoid muscle force.
Sx = 1840sin77 = 1790 N
Sy = 1840cos77 = 410 N
The Sy component is wrong, the answer key says that it should be -297 N downward... I am not sure how they are getting this answer??

 

[anathema]

Thank you for your post. I am a scientist and I would be happy to help you with your question.

First, let's review the given information. The bodybuilder is holding a dumbbell of weight Wd, while his arm weighs Wa = 31.0 N. The deltoid muscle is the only muscle acting and has a maximum force of M = 1840 N. We are given the distances that locate where the various forces act on the arm in Figure 9.8b.

To find the heaviest dumbbell that can be held, we need to consider the equilibrium of forces acting on the arm. In other words, the sum of all the forces acting on the arm must be equal to zero, since the arm is not moving. This can be represented mathematically as:

ΣF = 0

To find the heaviest dumbbell, we need to find the weight Wd that will cause the deltoid muscle force to reach its maximum value of 1840 N. This can be done by setting up a force balance equation in the vertical direction:

ΣFy = 0

This can be simplified to:

-Wd + Wa + Sy = 0

Substituting in the values given in the problem, we get:

-Wd + 31.0 + Sy = 0

Solving for Wd, we get:

Wd = 31.0 + Sy

Since we know that the maximum force the deltoid muscle can supply is 1840 N, we can set this equal to Sy:

1840 N = Sy

Therefore, the heaviest dumbbell that can be held is:

Wd = 31.0 + 1840 = 1871 N

To find the horizontal and vertical force components, Sx and Sy, we can use the trigonometric relationships between the angles and sides of a right triangle. In this case, we can use the angle of 77 degrees, which is the angle between the deltoid muscle and the horizontal.

Using the trigonometric functions, we can write:

Sx = M cos 77 = 1840 cos 77 = 410 N

Sy = M sin 77 = 1840 sin 77 = 1790 N

Therefore, the horizontal force component, Sx, is 410 N and the vertical force component, Sy, is 1790 N.

I hope this helps to clarify

 

[nlightner]

I would first clarify the problem and make sure that all the given information is correct. In this case, the weight of the arm (Wa) seems to be significantly larger than the weight of the dumbbell (Wd), which may not be realistic. I would also double check the given angles and distances to ensure accuracy.

Assuming all the given information is correct, I would approach this problem by analyzing the forces acting on the arm and shoulder joint. The deltoid muscle is the only muscle acting on the arm, so it must be providing a force equal and opposite to the weight of the arm and the weight of the dumbbell. This results in a total vertical force of 1840 N (31.0 N + Wd).

To find the horizontal force component (Sx), I would use the fact that the arm is extended horizontally and in equilibrium. This means that the torque (force x distance) acting on the arm must be zero. The torque due to the weight of the arm (Wa) would be counteracted by the torque due to the deltoid muscle force (1840 N) acting at a distance of 0.22 m. This results in the equation: Wa x 0.15 m = 1840 N x 0.22 m. Solving for Wa, we get Wa = 2440 N. This means that the horizontal force component (Sx) must be equal to 2440 N.

To find the vertical force component (Sy), I would use the fact that the arm is also in equilibrium in the vertical direction. This means that the sum of all the vertical forces must be equal to zero. Since we already know that the deltoid muscle is providing a vertical force of 1840 N, the vertical component of the weight of the arm (Wa) must be equal to -1840 N. This means that the vertical force component (Sy) must be equal to -1840 N.

In conclusion, the heaviest dumbbell that can be held is 86.1 N, the horizontal force component (Sx) is 2440 N, and the vertical force component (Sy) is -1840 N. These values indicate that the shoulder joint is providing a net force of 2440 N horizontally and -1840 N vertically to keep the arm in equilibrium.