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Mechanics M1 Question Help



A parcel A of mass 2kg rests on a rough slope inclined at angle θ\theta to the horizontal, where tan(θ)=0.75tan(\theta) = 0.75.
A string is attached to A and passes over a small smooth pulley at P. The other end of the string is attached to a weight B of mass 2.2kg, which hangs freely as shown above. The parcel A is in limiting equilibrium and about to slide up the slope. By modelling A and B as particles and the string as light and inextensible, find:

a) the normal contact force acting on A
b) The coefficient of friction between A and the slope


Thanks for any help
Start by drawing a diagram showing all the forces acting on A. Basic trig will give you the forces you don't get given explicitly in the question. Remember A is in equilibrium so the forces must be balanced.
Reply 2
You can find out theta from tanx=0.75.
You can find out the tension in the string because by finding the object B's weight (9.8m)
If it's in limiting equilibrium, about to move up the slope, which way is friction acting? Then what is the equation for friction?
Reply 3
mcp2
You can find out theta from tanx=0.75.


Is that the idea though? Are they not after a Sin/Cos ?
Reply 4
Mitch92uK
Is that the idea though? Are they not after a Sin/Cos ?
mcp2
You can find out theta from tanx=0.75.
You can find out the tension in the string because by finding the object B's weight (9.8m)
If it's in limiting equilibrium, about to move up the slope, which way is friction acting? Then what is the equation for friction?
Potally_Tissed
Start by drawing a diagram showing all the forces acting on A. Basic trig will give you the forces you don't get given explicitly in the question. Remember A is in equilibrium so the forces must be balanced.




not sure what the question means by "the normal contact force acting on A" !
Reply 5
Once you have that and you know which way gravity acts on object A and its mass, then you can find the normal contact force and force acting down the slope due to gravity.

Normal contact force is the force of the slope on object A
Reply 6
i get a 15.68 weight force and a 21.56 down slope force from the object B
can anyone tell me if this is correct, and what should i do next? just minus them?
Reply 7
21.56 is up the slope, it's the tension of the string.

Yes that is the correct normal contact force.
Reply 8
So should i take that and minus the force of weight through the slope which is 2gcos(theta) = 15.68?
Reply 9
Yeah, then what you have left is friction, and you know the normal contact force, so it's easy peasy to find the coeff of friction.
Milan.
i get a 15.68 weight force and a 21.56 down slope force from the object B
can anyone tell me if this is correct, and what should i do next? just minus them?


I agree 15.68N. I get mu equal to 0.625
Reply 11
steve2005
I agree 15.68N. I get mu equal to 0.625

i've sat here for half an hour trying to get that, i have no idea how to do it :frown:
i suck at mechanics
can you help me please!
Milan.
i've sat here for half an hour trying to get that, i have no idea how to do it :frown:
i suck at mechanics
can you help me please!


Are my answers correct?
Reply 13
steve2005
Are my answers correct?

I dont know i dont have answers :frown:
thats the only problem, otherwise i would just work towards answers/backwards
Milan.
I dont know i dont have answers :frown:
thats the only problem, otherwise i would just work towards answers/backwards


Reply 15
Milan.
So should i take that and minus the force of weight through the slope which is 2gcos(theta) = 15.68?


Do that then what you have left is Friction. You know the normal contact force. F=muN, where F=friction, mu=coeff of friction and N=normal contact force.