What smallest value of kinetic friction

AI Thread Summary
The discussion focuses on a physics problem involving a pulley system with two masses, m1 and m2, where m1 is on an inclined plane and m2 hangs vertically. The coefficient of kinetic friction is given as 0.15, and the goal is to determine the acceleration of the system and the minimum kinetic friction required to prevent acceleration. The user successfully calculated the acceleration as 2.16 m/s² for part A but struggles with part B, seeking clarification on how to set up the equations correctly to find the smallest value of kinetic friction. Participants emphasize the need for a clear symbolic representation of the equations to assist in solving part B. The conversation highlights the complexity of applying Newton's laws in systems with friction and inclined planes.
fizzex
Messages
4
Reaction score
0

Homework Statement


(a) suppose the coefficient of μκ between m1 and the plane in Fig 4-57 is 0.15, and that m1=m2=2.7 kg. As m2 moves down, determine the magnitude of the acceleration of m1 and m2, given θ = 25°. (b) What smallest value of kinetic friction will keep this system from accelerating?


Homework Equations


There are many equations...
For part A, ƩF = Fg - Ft +Fn
Fg = mg
ƩF = Fg - Ft
Ff = μ(Fn)
Part B, I'm not too sure.
I was going to have Ft - Fg - Ff = ƩF, and have ƩF=0 because ƩF = ma, and acceleration would be 0, but then I was confused.


The Attempt at a Solution


I solved Part A... the acceleration is 2.16. If you want me to go through the whole process, I will, just for the sake of time, I was hoping to just start from part B, where I'm absolutely stuck. I know that I have to set acceleration equal to zero, but when I tried I got 0.14 instead of the answer, 0.64.

Any suggestions? ...and does that even make sense? Sorry, it's my first post.
 
Physics news on Phys.org


Welcome to PH Forums.

It's hard to help without the figure!
 


I can't copy it, but it's a pulley system. M2 is hanging over the side of the inclined plane, and M1 is sliding down the 25 degree angle. Does that help?
 


I think you'll have to spell out your final equation for the acceleration (symbolic form, no numbers) so we can see where things stand.
 


From Part A, there were two equations involving acceleration (ƩF = ma)... ƩF = Fg - Ft and ƩF = Ft - Fg - Ff. The first one was applied to the hanging box, box 2, because there was no friction. (Conversely, the second equation included Ff, as box 1 was sliding down the inclined plane.)

These formulas were then set equal to each other, which found the answer to part A, finding the acceleration.

I'm not sure how to put them into the second part of the problem.
 
Thread 'Collision of a bullet on a rod-string system: query'
In this question, I have a question. I am NOT trying to solve it, but it is just a conceptual question. Consider the point on the rod, which connects the string and the rod. My question: just before and after the collision, is ANGULAR momentum CONSERVED about this point? Lets call the point which connects the string and rod as P. Why am I asking this? : it is clear from the scenario that the point of concern, which connects the string and the rod, moves in a circular path due to the string...
Thread 'A cylinder connected to a hanged 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...

Similar threads

Replies
19
Views
12K
Replies
3
Views
1K
Replies
6
Views
2K
Replies
2
Views
4K
Replies
25
Views
3K
Replies
4
Views
3K
Back
Top