Rotational dynamics and static equilibrium question

Click For Summary
The discussion revolves around a physics problem involving rotational dynamics and static equilibrium, specifically analyzing the forces acting on a crossing guard holding a STOP sign. Key points include the weights of the upper arm, lower arm, hand, and sign, as well as the forces exerted by the deltoid muscle and shoulder joint. Participants are encouraged to apply the three fundamental equations of static equilibrium: the sum of torques, the sum of forces in the x-direction, and the sum of forces in the y-direction. The initial query asks whether the force exerted by the deltoid muscle is greater than, less than, or equal to the horizontal component of the shoulder joint force. The discussion aims to guide users through solving for the various forces involved in this scenario.
hkstyles
Messages
2
Reaction score
0
Please help with this question. I don't know how to approach this difficult question

a crossing guard holds a STOP sign at arm's length, as shown. Her arm is horizontal, and we assume that the deltoid muscle is the only muscle supporting her arm. The wright of her upper arm is Wu = 18 N, the weight of her lower arm is Wl = 11 N, the weight of her hand is Wh = 4.0N, and the weight of the sign is Ws = 8.9 N. The location where each of these forces acts on the arm is indicated in the figure. A force of magnitude fd is exerted on the humerus by the deltoid, and the shoulder joint exerts a foce on the humerus with horizontal and vertical components given by fx and fy , respectively. (a) is the magnitude of fd greater than, less than, or equal to the magnitude of fx ? why?. Find (b) fd (c) fx and (d) fy

http://answerboard.cramster.com/Answer-Board/Image/2007115234686332990316848737506341.jpg
 
Last edited by a moderator:
Physics news on Phys.org
Think about the 3 equations involved...

sum of torques about any point = 0 (1)

sum of forces in x-direction = 0 (2)

sum of forces in y-direction = 0 (3)

equation (2) should let you answer a).

equation (1) gives you part b). try to go on from there to get the other parts...
 
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 18 ·
Replies
18
Views
2K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
Replies
8
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
  • · Replies 3 ·
Replies
3
Views
13K
  • · Replies 5 ·
Replies
5
Views
5K
  • · Replies 2 ·
Replies
2
Views
4K
Replies
1
Views
29K
  • · Replies 28 ·
Replies
28
Views
4K