Determine the force P required to move the block A of 5000 N weight up the inclined plane, coefficient of friction between all contact surfaces is 0.25. Neglect the weight of the wedge and the wedge angle is 15 degrees.
Given : Weight of block A = 5000 N
Wedge angle = 15º
To find : Force P required to move block A up the inclined plane
Refer to figure.If the co-efficient of friction is 0.60 for all contact surfaces and θ = 30o,what force P applied to the block B acting down and parallel to the incline will start motion and what will be the tension in the cord parallel to inclined plane attached to A.
Take WA=120 N and WB=200 N.
Two blocks A and B are resting against the wall and floor as shown in the figure.Find the minimum value of P that will hold the system in equilibrium. Take μ=0.25 at the floor,μ=0.3 at the wall and μ=0.2 between the blocks.
Given : μ=0.25 at floor
μ=0.2 between blocks
To find : Minimum value of force P