Equilibrium when applied to muscles and joints Help, please

Click For Summary
The discussion focuses on calculating the forces acting on the arm when held outstretched. The user successfully calculated the force required by the deltoid muscle (FM) to be 264 N. They were initially confused about finding the force exerted by the shoulder joint (FJ) but received guidance on using torque and force components to solve the problem. Ultimately, they determined that the joint force (FJ) is 260 N. The conversation highlights the application of equilibrium principles in analyzing muscle and joint forces.
confusedbyphysics
Messages
61
Reaction score
0
This seems like it should be easy but I am not sure what to do. Heres the problem:
The picture to the problem is here: http://www.webassign.net/giancoli/9-65alt.gif
"(a) Calculate the magnitude of the force FM required of the "deltoid" muscle to hold up the outstretched arm shown in Fig. 9-65. The total mass of the arm is 2.8 kg and = 12°.
(b) Calculate the magnitude of the force FJ exerted by the shoulder joint on the upper arm.
I found a) by doing .12 m (sin 12) F - .25 m (2.8 kg) (9.8 m/s^2) = 0. F = 264 N.
But how do I find the force FJ? Confused (we haven't had the lesson on this yet, but trying to get my homework done early because I have a busy week). THANKS for any help!
 
Physics news on Phys.org
your torque looks okay.

Now that you know all the Forces, you could treat it like a chapter 3 problem
(Sum of the force components (x,y) = 0 . you DID enjoy that, right?)

OR, you can sum torques around the tendon attachment point, to = 0
(pretending that it might rotate around this attachment point)
This gives the vertical component of the joint Force (Fjx applies no torque).
 
thanks lightgrav, adding up the components worked. answer turned out to be 260 N. Thank you for the help!
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

Similar threads

  • · Replies 1 ·
Replies
1
Views
16K
  • · Replies 3 ·
Replies
3
Views
10K
Replies
6
Views
15K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 1 ·
Replies
1
Views
985
Replies
6
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 5 ·
Replies
5
Views
14K
  • · Replies 5 ·
Replies
5
Views
3K