Calculate resultants of two forces

  • Thread starter Thread starter zeralda21
  • Start date Start date
  • Tags Tags
    Forces
AI Thread Summary
The discussion focuses on calculating the resultant of two vertical forces acting on a beam and determining where this resultant force should be applied. The calculated resultant force is 20N directed downward, with the x-component being zero. Participants express confusion about how the distances of the forces from point A affect the resultant's application point. It is noted that each force creates a moment about point A, which is crucial for understanding the system's equilibrium. The conversation emphasizes the need to consider both the magnitude and the location of the resultant force to maintain balance.
zeralda21
Messages
119
Reaction score
1

Homework Statement



Where does the resultant of the two forces act?

http://i.imgur.com/CBO8fi2.jpg

Homework Equations



None really. Basic algebra for calculating resultants.

The Attempt at a Solution



First off, I have tried to calculate the x and y components of the resultant, call it R. Here I think that R_{x}=0 (both forces are vertical) and R_{y}=-20 (net force in the y-direction). So the resultant is:

R=R_{x}+R_(y} = 0x-20y with magnitude of 20N. However, I am unable to understand how the lengths will help me.
 
Last edited:
Physics news on Phys.org
The question wants you to replace the two forces at different distances with one force that has the same effect. So - what is the effect of each force as drawn? (Hint: what is special about point A?)
 
Simon Bridge said:
The question wants you to replace the two forces at different distances with one force that has the same effect. So - what is the effect of each force as drawn? (Hint: what is special about point A?)

That is what I have tried to do actually. I have replaced them with one downward force with magnitude 20N. But I don't know where it is located.

What is special about point A? Well, not sure. There must be a force to compensate the downward force but that is done by the beam(object).
 
20N is what you'd get if the forces acted along the same line.
Do they?

Won't each of the two forces create a moment about point A?
 
Simon Bridge said:
20N is what you'd get if the forces acted along the same line.
Do they?

Won't each of the two forces create a moment about point A?

Yes, I suppose. I will have to think a little bit about this one before I answer. Not sure about the theory yet.
 
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