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# Mechanics AS question help! watch

1. In this question take g to be 10 m/s^2. A particle of mass 0.8kg is at rest on a rough horizontal plane. The coefficient of friction between the particle and the plane is 0.5 Find the least force required to pull the particle along the plane if the force is at an angle of 30 degrees to the plane.

i tried and tried but this is just....not working
help!
2. So you can find F using F=mu R (Make sure you resolve the forces)

Then you can Resolve the force 30 degrees and equate them together.
3. (Original post by 2710)
So you can find F using F=mu R (Make sure you resolve the forces)

Then you can Resolve the force 30 degrees and equate them together.

mm yep i think i got it
cheers
4. (Original post by 2710)
So you can find F using F=mu R (Make sure you resolve the forces)

Then you can Resolve the force 30 degrees and equate them together.

the part that i've wrote in the thread is part b). i got that now.

but in part a) it says all same but the ending is "find the leas force required to pull the particle along the plane if the force is horizontal.

which means that force W acts downwards and the other force acts horizontally to the right. what i dont understand is how you resolve them then?? just tell me the method, and i'll do it.

thank you so much
5. Well, then its already resolved...

So You can find F. And the you just equate that with the only horizontal force.
6. but that force that acts horizontally right is unknown the only force that is known is W.
i dont get it
7. (Original post by 2710)

So You can find F. And the you just equate that with the only horizontal force.

but that force that acts horizontally right is unknown the only force that is known is W.
i dont get it
8. (Original post by Boomka)
but that force that acts horizontally right is unknown the only force that is known is W.
i dont get it
Well its supposed to be unknown...that's what you are trying to find right?

F = mu R, you can find F (which is the force opposing the unknown force)

And in this case, F is also < Unknown force
9. (Original post by 2710)
Well its supposed to be unknown...that's what you are trying to find right?

F = mu R, you can find F (which is the force opposing the unknown force)

And in this case, F is also < Unknown force

ok...call me stupid but
coefficient of friction is 0.5
in this case you get F=0.5R
two unknowns !! you cant solve it. i dont see how you do anyway.
10. (Original post by Boomka)
ok...call me stupid but
coefficient of friction is 0.5
in this case you get F=0.5R
two unknowns !! you cant solve it. i dont see how you do anyway.
Lol. R = 0.8 x 10

Right?
11. (Original post by 2710)
Lol. R = 0.8 x 10

Right?
No

Let P be pulling force then

12. (Original post by 2710)
Lol. R = 0.8 x 10

Right?
but doesnt this 0.8 x 10 give you the weight?
because it says to take g as 10,
W=mg so W=0.8 x 10!
which is 8N, i get that.
but what about the R that acts horizontally to the right and we know nothing about?

sorry i seriously dislike my mechanics teacher in school.
13. (Original post by steve2005)
No

Let P be pulling force then

(Original post by Boomka)
but doesnt this 0.8 x 10 give you the weight?
because it says to take g as 10,
W=mg so W=0.8 x 10!
which is 8N, i get that.
but what about the R that acts horizontally to the right and we know nothing about?

sorry i seriously dislike my mechanics teacher in school.
Firstly. R does not act horizontally to the right, it acts perpendicular to the horizontal plane. R is equal and opposite to W, since those 2 are the only vertical forces. R is the reaction force of the plane on the object.

If R was 6N and W was 5N, (ie they were not the same) then the block will have to be moving Upwards. But we know it is in equilibrium vertical wise. Therefore all vertical components must equal each other

R = W

R = mg = 80

F = mu R

Find F

Now you know that the friction is a certain value. So what is the least you need to move it? It will be 0.000000000000......1N more

So unknown Force has to be > F. So F is the least force needed...kinda

Hope this helps
14. the weight is R because it is in equilibrium vertically right? so then you only need (pulling force)xsin(30) > F(=0.5x8x10).
15. (Original post by Harry-AA)
the weight is R because it is in equilibrium vertically right? so then you only need (pulling force)xsin(30) > F(=0.5x8x10).
This question has already been done. But anyways, this is wrong. R is not 0.8 x 10 in this question, since you need to resolve the unknown force as well.

And also, why would it be sin 30? More like cos 30?

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