Static equilibrium and torque problem

[physicshurts]

Homework Statement

The figure shows an outstretched arm with a mass (m1) of 4.4 kg. The arm is 56 cm long, and its center of gravity is 21 cm from the shoulder. The hand at the end of the arm holds a 6.0-kg mass (m2). (so its a man holding a dumbell, and his arm is 15 degrees up from the horizontal)

What is the torque about the shoulder due to the weights of the arm and the 6.0-kg mass?

If the arm is held in equilibrium by the deltoid muscle, whose force on the arm acts 5.0 below the horizontal at a point 18 from the shoulder joint what is the force exerted by the muscle?

The Attempt at a Solution

So, I got the first question by determining the force perpendicular to the arm: m2gcos15 and then multiplying that by the distance from the shoulder, .56m. Then I added (m1g)(.21m), the mass of the weight of the arm and the position of its center of mass (.21m)

Now the second question, I am really confused about. I know that I make the deltoid muscle into a tension force, but how do I find out what the tension force is, do I simply do a sum of force in the x and y direction? Or do I have to include torque some how?

 

[LowlyPion]

Now the second question, I am really confused about. I know that I make the deltoid muscle into a tension force, but how do I find out what the tension force is, do I simply do a sum of force in the x and y direction? Or do I have to include torque some how?

Welcome to PF.

Apparently you have a 5 degree angle made by the muscle and at an attached distance.

Sounds like you have a triangle there, where the vertical component of the force will need to support the weight of the arm.

 

[physicshurts]

Im still not sure how to set it up though? How do I find the force exerted by the muscle? Do I need to find torque again?

 

[LowlyPion]

Do I need to find torque again?

Yes.

Figure the point that the weight is acting through (CofM) times the distance and that equals the vertical component of the force exerted by the muscle. The angle is tiny, so the force will be large right? And it acts through its attachment point on the arm.

 

[geekoid32]

Hi there! I have used the physics forum as a resource before although I must say this is the first time I have ever been "stuck" enough to ask directly.

I can't seem to set up the problem correctly. I know that the deltoid muscle works in an upward force and that its Force = .18 * sin (5 degrees). However, I don't quite understand what to set it equal to (since the torques equal zero) -- help!

 

[LowlyPion]

Hi there! I have used the physics forum as a resource before although I must say this is the first time I have ever been "stuck" enough to ask directly.

I can't seem to set up the problem correctly. I know that the deltoid muscle works in an upward force and that its Force = .18 * sin (5 degrees). However, I don't quite understand what to set it equal to (since the torques equal zero) -- help!

Welcome to PF.

If I understand the drawing - which I am not advantaged in being able to see I might note - the .18 distance is the lever over which it acts. The sum of the torques being 0, means that the weight acting at the center of mass, whatever that is, as well as the weight held in the hand, will equal the Tension in the muscle/sin5° and that is all times the .18 distance.