1 Answer | Add Yours
For a block placed on an inclined ramp which is not frictionless, there are two forces acting. One of them is the gravitational force of attraction that makes the block slide down the ramp and a counteracting force due to friction.
The ramp in the problem is inclined at an angle of 30 degrees. The force with which it is pulled down due to gravity is equal to the product of its mass m and the acceleration due to gravity g.
We divide this force F = mg, into two components, one along the ramp in the downward direction which is equal to m*g* cos 60 and one perpendicular to the ramp equal to m*g* sin 60.
The force of friction opposing the motion of the block is Mu * N, where Mu is the coefficient of friction and N is the normal force.
Here the opposing force is .65*m*g* sin 60. This has to be less than the component of the force moving the block downward.
This gives 0.65*m*g* sin 60 =< m*g* cos 60
We see that m and g cancel from both the sides and we are left with
=> 0.65 sin 60 =< cos 60
This implies that whether the block slides down or not is not decided by its mass, but by the inclination of the slope and the coefficient of friction. Here 0.65* sin 60 = .5629 is greater than cos 60 = 0.5
Therefore the block does not slide down the ramp irrespective of its mass.
We’ve answered 319,859 questions. We can answer yours, too.Ask a question