Finding the Third Force to Achieve Zero Total Torque on an Equilateral Triangle

  • Thread starter Thread starter merced
  • Start date Start date
  • Tags Tags
    Torque Triangle
AI Thread Summary
To achieve zero total torque on an equilateral triangle with forces F_1 and F_2 acting along two sides, a third force F_3 must be applied at point B along side BC. The sum of the torques around point O must equal zero, necessitating the calculation of the forces in component form. Identifying the perpendicular components of the forces relative to point O is crucial for determining the required force F_3. Participants suggest preparing calculations and visual aids to clarify the problem. Understanding the relationship between the forces and their points of application is essential for solving the torque equation effectively.
merced
Messages
44
Reaction score
1
http://img300.imageshack.us/img300/4792/physics1gj9.th.png

"Two forces F_1 and F_2 act along the two sides of an equilateral triangle as shown...Point O is the intersection of the altitudes of the triangle. Find a third force F_3 to be applied at B and along BC that will make the total torque zero about the point O."

So the sum of the torques = 0. I don't know what to do first. Should I put the forces into component form and take the tangential component of the forces??
 
Last edited by a moderator:
Physics news on Phys.org
Unfortunately, I can't see the image, but putting the forces into components would be a good start.
 
Click on the image.
 
merced said:
Click on the image.

I don't have anything to click on, it's my computer, probably. Someone else will be able to give an exact reply to you. In the meantime, prepare some work to present.
 
I guess what I mean is, how do I determine what is perpendicular to the intersection (O)? Because I have to use the perpendicular force right?
 

Thread 'Errors and significant figures'

I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
View full post »
Thread 'A cylinder connected to a hanging mass'

Thread 'A cylinder connected to a hanging mass'

Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
View full post »

Similar threads

Trivial question regarding Newton's Third Law and Elasticity
Replies
17
Views
2K
To find the resultant of forces on a lamina
Replies
5
Views
3K
To find the ratio of forces around a right triangle
Replies
2
Views
2K
To find the magnitude of the resultant of forces around a triangle
Replies
4
Views
2K
Calculating Net Gravitational Force in Equilateral Triangle Configuration
Replies
5
Views
2K
An equilateral triangle's electric field at its center
Replies
10
Views
4K
Finding the Point of Zero Electric Field in an Equilateral Triangle of Charges
Replies
12
Views
2K
Irodov Equilateral Triangle Problem
Replies
6
Views
9K
Why Isn't the Sum of Torques Zero in This Ladder Equilibrium Problem?
Replies
8
Views
6K
Can we find the frequency of a rod pendulum by just using F=ma?
Replies
7
Views
2K

Hot Threads

Recent Insights

Back
Top