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A rectangular box lies on a rough inclin...

A rectangular box lies on a rough inclined surface The co-efficient of friction beteen the surface and the box is `mu`. Le the mass of the box ne m What is the force acting on the box down the plane if the angle of inclination of the plane is increased to `alpha gt theta` ? .

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Force acting on the box down the plane
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A rectangular box lies on a rough inclined surface . The coefficient of friction between the surface and the box is mu . Let the mass of the box be m . (a) At what angle of inclination theta of the plane to the horizontal will the box just start to slide down the plane ? (b) What is the force acting on the box down the plane , if the angle of inclination of the plane is increased to alpha gt theta ? (c) What is the force needed to be applied upwards along the plane to make the box either remain stationary or just move up with uniform speed ? (d) What is the force needed to be applied upwards along the plane to make the box move up the plane with acceleration a ?

A block of mass m rests on a rough inclined plane. The coefficient of friction between the surface and the block is µ. At what angle of inclination theta of the plane to the horizontal will the block just start to slide down the plane?

A block of mass m is placed on a rough inclined plane. The corfficient of friction between the block and the plane is mu and the inclination of the plane is theta .Initially theta=0 and the block will remain stationary on the plane. Now the inclination theta is gradually increased . The block presses theinclined plane with a force mgcostheta . So welding strength between the block and inclined is mumgcostheta , and the pulling forces is mgsintheta . As soon as the pulling force is greater than the welding strength, the welding breaks and the blocks starts sliding, the angle theta for which the block start sliding is called angle of repose (lamda) . During the contact, two contact forces are acting between the block and the inclined plane. The pressing reaction (Normal reaction) and the shear reaction (frictional force). The net contact force will be resultant of both. Answer the following questions based on above comprehension: Q. If mu=(3)/(4) then what will be frictional force (shear force) acting between the block and inclined plane when theta=30^@ :

A body of mass M is applying horizontal force to slide a box of mass M_(1) on a rough horizontal surface The coefficient of friction between the shoe of the boy and the floor is mu and that between the box and the floor is 'mu_(1)' In which of the following cases is it certainly not possible to slide the box ? .

A block of mass m slides down an inclined plane of inclination theta with uniform speed The coefficient of friction between the block and the plane is mu . The contact force between the block and the plane is .

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