Why Am I Getting a Positive Force of Friction Instead of Negative?

  • Thread starter Thread starter HelloImAsh
  • Start date Start date
  • Tags Tags
    Friction
AI Thread Summary
The discussion centers on a physics problem involving the calculation of the force of friction between two materials, where the coefficient of sliding friction is 0.35 and the mass of the object is 5.0 kg. The calculated force of friction is 17.15 N, but the user is confused about why the result is positive instead of negative, given their definition of forward as positive. It is clarified that the equation F1=μF2 provides only the magnitude of friction, which is inherently a scalar quantity, while the direction of the force of friction opposes the motion. Thus, the confusion arises from a misunderstanding of how to represent the direction of the force in vector terms. Understanding that friction acts opposite to the direction of motion resolves the issue of sign in the calculation.
HelloImAsh
Messages
6
Reaction score
0

Homework Statement


The coefficient of sliding friction between two materials is 0.35. A 5.0 kg object made of one material is being pulled along a table made of another material. What is the force of friction?

I'm fairly certain I solved the problem correctly, the thing is, I should be getting a negative force as the answer (because I defined forward as positive), and I'm getting a positive force. Could someone please point out which part of my solution is incorrect.

Homework Equations


F1 is force of sliding friction, F2 is normal force, and μ is the coefficient of sliding friction:
F1=μF2

The Attempt at a Solution


F1=μF2
=0.35(mg)
=17.15N
 
Physics news on Phys.org
HelloImAsh said:

Homework Statement


The coefficient of sliding friction between two materials is 0.35. A 5.0 kg object made of one material is being pulled along a table made of another material. What is the force of friction?

I'm fairly certain I solved the problem correctly, the thing is, I should be getting a negative force as the answer (because I defined forward as positive), and I'm getting a positive force. Could someone please point out which part of my solution is incorrect.

Homework Equations


F1 is force of sliding friction, F2 is normal force, and μ is the coefficient of sliding friction:
F1=μF2

The Attempt at a Solution


F1=μF2
=0.35(mg)
=17.15N
The equation F1=μF2 only gives the magnitude of the frictional force. The force of friction, F1, is in a direction perpendicular to the normal force, F2 , so that equation cannot be true in a vector sense.

The force of friction is in such a direction as to oppose the relative motion of the two objects.
 
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