How Do You Calculate the Third Force in a Three-Force System?

  • Thread starter Thread starter Callen9
  • Start date Start date
  • Tags Tags
    Homework Vector
AI Thread Summary
To calculate the third force in a three-force system, apply Newton's second law, F = ma, using the known forces and acceleration. The object has a mass of 20 kg and experiences an acceleration of -8 m/s² in the i direction and 6.0 m/s² in the j direction. Two forces, F1 and F2, are given, and the net force can be expressed as F1 + F2 + F3 = ma. Separate equations for the x and y components should be established to solve for the unknown force F3. By substituting the known values and solving the equations, the third force can be determined accurately.
Callen9
Messages
5
Reaction score
0

Homework Statement



A 20-kg object is subjected to three forces which produce an acceleration a = -8 m.s-2 i + 6.0 m.s-2 j on the object. Two of the forces are:
F1 = 3.0 N i + 16.0 N j
F2 = -12.0 N i+ 8.0 N j
Find the third force.



Homework Equations





The Attempt at a Solution


To be honest, I've been doing physics for the past 8 hours so I'm brain dead right now. This is my last problem on my homework. I know that F=ma, so I'm thinking maybe I have to set up an equation like (F1+F2+F3)=20a. But something about that doesn't seem right. Any help and I would be very grateful.

Thanks,
 
Physics news on Phys.org
You're exactly right. Use F=ma and set up two equations for the two components. Set up one equation for the i direction and one equation for the j direction. F1+F2+F3=ma, then you just need to solve for F3
 
Write separate equations for the vector components of the equation F = m*A. Here F is the net force acting on mass m, so F = F1 + F2 + F3. each of these F's have two vector components (x,y directions).

So suppose you start by looking at just the x-components. Can you write the F = m*A equation for that?
 
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...
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...
Back
Top