Size of the friction force felt?

AI Thread Summary
The discussion revolves around calculating the friction force on a sled being pulled at a constant velocity. The sled is pulled with a force of 142 N at a 32-degree angle, and its weight is 119 N. The normal force was initially miscalculated but corrected to approximately 43.75 N after considering the vertical component of the pulling force. Participants emphasize the importance of using Newton's First Law and suggest drawing a free body diagram to visualize the forces acting on the sled. The friction force must equal the horizontal pulling force for the sled to maintain constant velocity.
EricMatthew
Messages
6
Reaction score
0
Consider a sled being pulled across the snow at a constant velocity. Imagine that a child pulls the sled with a force of 142 N, and that the rope makes an angle of 32 degrees with the horizontal. The sled moves along the snow with a constant velocity. The sled has a weight of 119 N.
a)Draw a diagram showing all the forces exerted on the sled. Break the forces up into components if needed, draw those forces, and come up with equations for those forces.b)What is the size of the normal force felt by the sled?

c)What is the size of the friction force felt by the sled?

The Attempt at a Solution


I'm having trouble finding the answer to c and I'm lost as to where to get started.

F1 = 142 N
F1y = 142sin(32)
F1x = 142cos(32)

b. ) Normal force is 119 - 142sin(32) = 118.4

b.) All the tutorials I found online mention two different formulas for kinetic and static and the nature of the question leaves me scratching my head as to which one I should apply. Should I just simply divide the weight of the sled by the normal force as such 119/118.4 since it is going at a constant velocity?
 
Physics news on Phys.org
EricMatthew said:
Consider a sled being pulled across the snow at a constant velocity. Imagine that a child pulls the sled with a force of 142 N, and that the rope makes an angle of 32 degrees with the horizontal. The sled moves along the snow with a constant velocity. The sled has a weight of 119 N.
a)Draw a diagram showing all the forces exerted on the sled. Break the forces up into components if needed, draw those forces, and come up with equations for those forces.b)What is the size of the normal force felt by the sled?

c)What is the size of the friction force felt by the sled?

The Attempt at a Solution


I'm having trouble finding the answer to c and I'm lost as to where to get started.

F1 = 142 N
F1y = 142sin(32)
F1x = 142cos(32)

b. ) Normal force is 119 - 142sin(32) = 118.4

b.) All the tutorials I found online mention two different formulas for kinetic and static and the nature of the question leaves me scratching my head as to which one I should apply. Should I just simply divide the weight of the sled by the normal force as such 119/118.4 since it is going at a constant velocity?

You should check your arithmetic in the normal force calculation before proceeding ...
 
119 - 142sin (32) = 43.751

Whoops, thank you for that.
 
So what would be a good way to proceed with c.?
 
EricMatthew said:
So what would be a good way to proceed with c.?
When the sled if being pulled at constant velocity, what must the friction force be, given the amount of pulling force as described in the problem statement?

If you need to, draw a free body diagram of the sled, and remember Newton's First Law of Motion:

http://teachertech.rice.edu/Participants/louviere/Newton/law1.html
 
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