Newtons law, interacting objects question?

AI Thread Summary
The discussion revolves around a physics problem involving two blocks with different masses and friction coefficients. Participants emphasize the importance of drawing free body diagrams for each block to identify the forces acting on them. The static friction coefficient should be used to determine the force required to initiate movement, while the kinetic friction coefficient applies once the blocks start sliding. There is a focus on applying Newton's laws, specifically F=ma, to calculate the forces and accelerations involved. Understanding the distinction between static and kinetic friction is crucial for solving the problem accurately.
Xinio64
Messages
4
Reaction score
0
Newtons law, interacting objects question??

A block of mass 5.00 Kg sits on top of a second block of mass 15.0 Kg, which in
turn sits on a horizontal table. The coefficients of friction between the two blocks are
: μs = 0.300 and μk = 0.100. The coefficients of friction between the lower block and
the table surface are: μs = 0.500 and μk = 0.400. You apply a constant horizontal
force to the lower block so that this force is just large enough to make this block start
sliding out from between the upper block and the table.

(a) Draw the free body diagram for each block, naming the forces on each of them.
(b) Determine the magnitude of each force on each block at the instant you have
started pushing but motion has not started.
(c) Determine the acceleration you measure for each block.
 
Physics news on Phys.org
Welcome to PF!

Hi Xinio64! Welcome to PF! :smile:

You must tell us how far you've got, so that we know what's troubling you, and how to help.

What have you tried? have you drawn the diagram?

What equations have you got? :smile:
 
tiny-tim said:
Hi Xinio64! Welcome to PF! :smile:

You must tell us how far you've got, so that we know what's troubling you, and how to help.

What have you tried? have you drawn the diagram?

What equations have you got? :smile:

Well I drawn two free body diagrams, one representing each block. The upper block has a weight and normal force of of 49N and a static frictional force pointing to the right. The lower block has a weight and normal force of 196N and a static frictional force of 78.4N pointing to the left and a constant force F pointing to the right. I don't know if in the free body diagrams I use kinetic friction or static friction forces. Equations I guess is F=ma. That's all I have.
 
Xinio64 said:
I don't know if in the free body diagrams I use kinetic friction or static friction forces. Equations I guess is F=ma. That's all I have.

Hi Xinio64! :smile:

To get the starting force, you must use the static coefficient.

To get the movement, you use the kinetic coefficient (but of course you use the force you got from the static coefficient).

So … what value force do you get, and how do you carry on from there? :smile:
 
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