What force does the dynamometer display?

  • Thread starter Thread starter Kamataat
  • Start date Start date
  • Tags Tags
    Dynamometer Force
AI Thread Summary
When two people pull on a dynamometer with equal forces of 25N, the device displays 25N due to the balance of forces. If only one person were pulling, the dynamometer would show zero, as it would move towards the person pulling. This scenario illustrates Newton's third Law of Motion, where every action has an equal and opposite reaction. Understanding this principle clarifies why the dynamometer reads the force exerted by both individuals. The discussion emphasizes the importance of grasping fundamental physics concepts for accurate interpretation of measurements.
Kamataat
Messages
137
Reaction score
0
Hi! Really simple question:

Two people are pulling on the ends of a dynamometer, each with a force of 25N. What force does the dynamometer display?

Textbook answer: 25N

I'm having a difficult time understanding this conceptually. Could anybody explain why the answer is 25N?

Thanks,
Kamataat
 
Physics news on Phys.org
If the other person wasn't pulling back, the dynomometer, would simply move towards the one person pulling it. It would therefore record zero.
Having another person pulling back, is exactly the same as having one end secured to a wall/floor. Perhaps you can understand that idea better.

This is a question about understanding Newton's third Law of Motion. Do you know that?
 
Yes, it's clear now. Thanks!
 
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