Proving that net torque isn't reliant on point of rotation.

Click For Summary
SUMMARY

This discussion centers on the calculation of net torque for a horizontal bar with varying forces and distances from different points of rotation. Using the formula Torque = r * Force * sin(theta), the participant calculated net torques for two scenarios: one with the point of rotation at the edge of the bar and another shifted 0.25m to the right. The results showed that net torque values were 10.17 Nm and 4.5 Nm, respectively, demonstrating that the point of rotation does not affect the net torque when forces and distances are appropriately accounted for. The participant seeks clarification on the relationship between distances, angles, and forces, suggesting the need for a diagram for better understanding.

PREREQUISITES
  • Understanding of torque and its calculation using the formula Torque = r * Force * sin(theta)
  • Familiarity with basic trigonometry, particularly sine functions
  • Knowledge of forces and their application on rigid bodies
  • Concept of rotational dynamics and points of rotation
NEXT STEPS
  • Study the effects of varying angles on torque calculations in different scenarios
  • Learn about the principles of static equilibrium in rigid body mechanics
  • Explore the use of free-body diagrams to visualize forces and torques
  • Investigate the relationship between net torque and angular acceleration in rotational motion
USEFUL FOR

Physics students, educators, and anyone interested in understanding the principles of torque and rotational dynamics in mechanical systems.

P-Fry
Messages
1
Reaction score
0

Homework Statement


So we have a horizontal bar. Distance = r Forces = F
All numbers remain constant with the exception of the distance, denoted as r(set)()

Length of bar = 1m

F1 = 10N r1-1 = 0m r2-1 = .25m (behind the point of rotation)
F2 = 5N r1-2 = 0.5m r2-2 = .25m (ahead of the point of rotation)
F3 = 20N r1-3 = 1.0m r2-3 = .75m (ahead)
F4 = 40N r1-4 = 0.75m r2-4 = .50m (ahead)

Theta 1 = 30 degrees clockwise from horizontal (South of east)
Theta 2 = 90 degrees, perpendicular, counterclockwise from horizontal (North)
Theta 3 = 60 degrees counterclockwise from horizontal (North of east)
Theta 4 = 90 degrees perpendicular, clockwise from the horizontal (south)

Essentially, with the first problem, the point of rotation is at the very edge, on the left. The second, it's shifted to the right .25m.

Homework Equations


Torque = r*Force*sin(theta)

The Attempt at a Solution


T1-1 = 0*(10) * sin(30) = 0
T1-2 = .5 *(5) * sin(90) = 2.5 (ccw)
T1-3 = 1*(20) * sin(60) = 17.32 (ccw)
T1-4 = .75 * (40) * sin (90) = 30 (cw)

T1net = 0 - 2.5 - 17.32 + 30 = 10.17

T2-1 = .25 * (10) * sin(30) = 1.25 (ccw)
T2-2 = .25 *(5) * sin(90) = 1.25 (ccw)
T2-3 = .75 * (20) * sin(60) = 12.99 (ccw)
T2-4 = .50 * (40) * sin (90) = 20 (cw)

T2net = -1.25 - 1.25 - 12.99 + 20 = 4.5

I can see logically why point of rotation doesn't have an effect on Net Torque, but I think I'm doing something wrong with the calculations.
 
Physics news on Phys.org
You need to explain how all these distances and angles relate to the bar and the force. A diagram would be good.
 

Similar threads

  • · Replies 33 ·
2
Replies
33
Views
3K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 22 ·
Replies
22
Views
4K
  • · Replies 5 ·
Replies
5
Views
3K
  • · Replies 4 ·
Replies
4
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 7 ·
Replies
7
Views
3K
Replies
6
Views
2K
  • · Replies 44 ·
2
Replies
44
Views
7K