Mechanics Help
Stuck with part ii) and iii), I think I have done i) correctly but if not please tell if I'm wrong and why
With ii) and iii), I keep getting expressions and not actual values for the magnitude of limiting friction, coefficient of friction etc. Is there anything I'm missing here?
Thanks
The force up the slope is equal to the force down the slope. So the component of the weight of the block down the slope is equal to the friction force. Using that force, you can then work out the co efficient of friction in the next part.
Original post by Sandrey546
The force up the slope is equal to the force down the slope. So the component of the weight of the block down the slope is equal to the friction force. Using that force, you can then work out the co efficient of friction in the next part.
How do you know that the cube is going down the slope at a steady speed?
what level is this?
Original post by Gent2324
what level is this?
A level
Original post by Yatayyat
A level
which board? im doing aqa and the coefficient of friction is not in the spec
Original post by Gent2324
which board? im doing aqa and the coefficient of friction is not in the spec
I don't think its a question from an actual exam. My teacher just gave it to me for homework for practice
Original post by Yatayyat
How do you know that the cube is going down the slope at a steady speed?
It is not moving at a steady speed. It is not moving at all. If there is a frictional force then that means there is an equal and opposite force against friction. There is no resultant because it is not moving so they are equal
Original post by Sandrey546
It is not moving at a steady speed. It is not moving at all. If there is a frictional force then that means there is an equal and opposite force against friction. There is no resultant because it is not moving so they are equal
Ahh so simply put it the cube is in equilibrium since it's stationary (i.e. zero resultant force). And equal and opposite forces which are parallel to the slope are component of cube weight down the slope to frictional force between the cube and steel sheet up the slope?
That means the frictional force is equal to mgsin(theta) = 1*9.8*sin(40) = 6.3N
and
coefficient of friction is 'mgsin(theta)/mgcos(theta) = tan(theta) = tan(40) = 0.84
Would you agree?
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