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mechanics problem hELP PLS

hello everyone,

i seem to be able to resolve just fine when a particle is on a slope, but when it rests on a horizontal plane and the friction is limiting, i get a little confused. Any help would be much appreciated please :P

a block of mass 4kg rests on a rough horizontal table. When a force of 6N acts on the block at an angle of 30 to the horizontal in a downward direction, the block is at the point of slipping. Find the coefficient of friction between the block and the table.
Original post by Mimi9335
hello everyone,

i seem to be able to resolve just fine when a particle is on a slope, but when it rests on a horizontal plane and the friction is limiting, i get a little confused. Any help would be much appreciated please :P

a block of mass 4kg rests on a rough horizontal table. When a force of 6N acts on the block at an angle of 30 to the horizontal in a downward direction, the block is at the point of slipping. Find the coefficient of friction between the block and the table.


OK so resolve the main 6N force horizontally and vertically. There will be a μR\mu R friction force acting on the particle. We know that for a particle to remain stationary on some rough surface, we need that FμRF \leq \mu R where F is the resultant force. When it is limiting, we get that F=μRF = \mu R.
Reply 2
Original post by RDKGames
OK so resolve the main 6N force horizontally and vertically. There will be a μR\mu R friction force acting on the particle. We know that for a particle to remain stationary on some rough surface, we need that FμRF \leq \mu R where F is the resultant force. When it is limiting, we get that F=μRF = \mu R.


yep i see, ive resolved the 6N horizontally and vertically (getting 3√3 and 3, respectively). I didnt realise that the horizontal component of the force is also the value of the friction force. The it`s just a matter of rearranging, as you`ve stated

I got 0.123, which is the correct answer, tHANK YOU SO MUCH for your help!!
i really appreciate it :biggrin:

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