What is the Correct Angle for Tension Force in Arm Mechanics?

In summary, the conversation discusses solving for three different parts, with a focus on using the law of static equilibrium to determine the necessary x and y components of the forces. The result of part a is a tension force of 870N in the deltoid, while part b involves finding the x and y components for Fs using the deltoid tension force and the weight of the arm. The magnitude of Fs is then calculated using the Pythagorean theorem. For part c, using the x and y components from part b, an attempt is made to find the angle using arctan, but the result is incorrect. The conversation ends with a request for equations and clarification on the directions of the two forces.
  • #1
Spocktastic
1
0
Homework Statement
The arm in the figure below weighs 43.2 N. The force of gravity acting on the arm acts through point A. Assume that L1 = 0.0890m, L2 = 0.342m and alpha = 11.0deg.
a) Determine the magnitude of the tension force Ft in the deltoid muscle.
b) Determine the magnitude of the tension force Fs of the shoulder on the humerus (upper-arm bone) to hold the arm in the position shown.
c) Determine the angle of tension force Fs relative to the x-axis.
Relevant Equations
Static Equilibrium: Fnet = 0 & Torque net = 0
Torque = radius*sin(angle)*Force
a^2 + b^2 = c^2
arctan = Fsin(angle)/Fcos(angle)
Screenshot 2022-11-21 at 3.36.51 PM.png


I already solved for part a, setting the sum of the Torques of the arms and deltoid equal to 0 and subbing in values which lead to a tension force of 870N in the deltoid.

For part b, I remembered the law of static equilibrium, so the summation x and y components of all the forces in the system must equal 0. I found that the x component for Fs needs to be 854.019 by multiplying the deltoid tension force by cos(11) and the y component for Fs to be the sum of the weight of the arm and the deltoid tension force*sin(11), which was 209.2. I used Pythagorean theorum to find that the magnitude of Fs is 879.27N.

Now for part c, I tried using the x and y components I got for part b and substituted them into arctan, so arctan(209.2/854.019) which resulted in me getting 13.76deg. However, it is wrong even when I try to answer it as -13.76. Is it the x-y components I messed up on?
 
Physics news on Phys.org
  • #2
Hello @Spocktastic. Welcome to PF!

Please present your equations. Its better if we can see directly what you did to get your result. This generally makes it easier for anyone helping to review your work.

$$ \rightarrow^+ \sum F_x = ...$$

$$ \uparrow^+ \sum F_y = ...$$

$$ \circlearrowright^+ \sum M_O = ... $$

Please try to format them using Latex, it doesn't take long to learn.
 
Last edited:
  • #3
Spocktastic said:
the sum of the weight of the arm and the deltoid tension force*sin(11)
In which directions do those two forces act?
 

Related to What is the Correct Angle for Tension Force in Arm Mechanics?

What is tension force angle on an arm?

Tension force angle on an arm refers to the direction and magnitude of the force acting on an arm or limb. It is the angle at which the force is applied to the arm.

Why is it important to figure out tension force angle on an arm?

Knowing the tension force angle on an arm is important because it helps in understanding the mechanics of the arm and how it moves. It also helps in determining the amount of force needed to perform certain actions and can aid in injury prevention.

How is tension force angle on an arm calculated?

Tension force angle on an arm can be calculated using trigonometry. The angle can be determined by dividing the horizontal component of the force by the total force and taking the inverse cosine of the result.

What factors can affect tension force angle on an arm?

The tension force angle on an arm can be affected by the direction and magnitude of the applied force, the weight and length of the arm, and the position and movement of the arm.

How can tension force angle on an arm be measured?

Tension force angle on an arm can be measured using a force gauge or dynamometer. These devices can provide accurate readings of the force applied and the direction in which it is acting on the arm.

Similar threads

  • Introductory Physics Homework Help
Replies
18
Views
231
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
13
Views
2K
  • Introductory Physics Homework Help
Replies
19
Views
915
Replies
3
Views
227
  • Introductory Physics Homework Help
Replies
2
Views
6K
  • Introductory Physics Homework Help
Replies
7
Views
4K
  • Introductory Physics Homework Help
Replies
7
Views
1K
  • Introductory Physics Homework Help
Replies
12
Views
3K
Back
Top