What is the acceleration and tension in this inclined pulley system?

  • Thread starter Thread starter Brynhildur
  • Start date Start date
  • Tags Tags
    Acceleration Zero
Join the discussion
Registration is free. Start your own thread to ask a follow-up.
7 replies · 3K views
Brynhildur
Messages
7
Reaction score
1
I have an inclined pulley with inclination 30°. There is mass A on top of it equals 2 kg. And over a frictionless pulley there is a mass B with weight 0,83 kg.
The text says that the coefficient of friction is 0,18 between mass A and the incline.
I need to know the acceleration. and I also have to find the tension? in the string between the masses over the pulley.
I know the answer is zero, but I'm having trouble understanding why? Is it because the mass of object B is too small to move the system as in over come the friction?
The equations I'm using are
Fþverkraftur*=m*g*sin(θ)
Ffriction = μ*m*g*cos(θ)
Ftension - mB*g=mb*a

*This is icelandic, I don't know the english term for it, I´m sorry :)
 
Physics news on Phys.org
Brynhildur said:
I have an inclined pulley with inclination 30°. There is mass A on top of it equals 2 kg. And over a frictionless pulley there is a mass B with weight 0,83 kg.
The text says that the coefficient of friction is 0,18 between mass A and the incline.
I need to know the acceleration. and I also have to find the tension? in the string between the masses over the pulley.
I know the answer is zero, but I'm having trouble understanding why? Is it because the mass of object B is too small to move the system as in over come the friction?
The equations I'm using are
Fþverkraftur*=m*g*sin(θ)
Ffriction = μ*m*g*cos(θ)
Ftension - mB*g=mb*a

*This is icelandic, I don't know the english term for it, I´m sorry :)
You have to crank out the numbers to see if the block A on the incline moves down the incline , up the incline, or remains stationary. Assume motion one way or the other or assume stationary and draw free body diagrams of each block to see what happens when applying Newtons laws. Note that tension forces cannot be pushing forces. I also wish I could translate Icelandic for you.
 
First thing to determine: Does it move? Imagine the blocks starting from rest. How much of a force must the tension overcome to accelerate block A? What's the maximum value of tension in the string?
 
So I've done a calculation and I'm not really sure if it's what I am looking for, or really how to interpret it
Ffriction=μN
ΣFx=T-Ffriction-Fþverkraftur=0
ΣAFy=N-Fg⋅yA=0
Σ T-Fg⋅B= 0

FgA=mA*g*sin(θ)-mag*cos(θ)
FyB= mB*g

T-Ffriction-mA*g*sin(θ)=0
N-mAg*cos(θ)
T-mBg=0 → T=mBg

mb*g-Ffriction-mAg*sin(θ)=0
Ffriction=mBg-mAg*sin(θ)
N=mAg*cos(θ)

μN=μ*mAg*cos(θ)= Ffriction

__________________
1. Ffriction=g(mB-mA*sin(θ))= 9,8(0,83-2*sin(30))=-1,66 N
2. Ffmax = μ*mAg*cos(θ) = 0,18*2*9,8*cos(30)= 3,055

The first one is the frictional force needed to prevent the block from sliding. The second is the maximum frictional force. Since the first is less than the second that means the blocks do not move and that is why there is zero acceleration? And the tension in the string is equal to the gravitational pull of mass B? 8,134 N?

I hope this is understandable but I can't seem to even google what I want to know, well how to put it in english and icelandic sites don't give me any information, so am I on the right track at least? :)
 
You're on the right track, but your reasoning is a bit difficult to follow.

Answer this question. If there were no friction, which way would mass A accelerate? Up the incline or down the incline? That will help you determine the direction of the friction force that must be overcome.
 
Calculate the net force between the two blocks, is this greater than the friction force ? if not then no movement will take place.
 
Yes, that is exactly what I was looking for! Thank you all very much!
 
Brynhildur said:
þverkraftur*

*This is icelandic, I don't know the english term for it, I´m sorry :)
Google translate gives "transverse force", which is reasonable, but it would be clearer to say force parallel to the plane, or force down the plane.