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Mechanics question
A crate of mass 250kg is on a slope inclined 20 degrees to the horizontal.
Find the frictional force. (I have done this part and i got 838N)
The crate is pushed horizontally with a force of 2000N. Show the crate remains in equilibrium.
any help would be good thanks
i have added a diagram below if that helps
thanks

Given that the crate is in equilibrium (before the horizontal force is applied), you know that the coefficient of friction must be greater than $\displaystyle \frac{838}{250g\cos20^\circ}\,N$.
After the horizontal force is applied, the forces acting on the crate are 200N horizontally, 250gN vertically, and a frictional force which will be acting down the slope. Resolve in the direction of the slope to find how large the frictional force must be for the system to remain in equilibrium. Then resolve in the perpendicular direction to find the normal force on the plane. If the ratio of those two forces is less than the coefficient of friction then you can conclude that the crate will not slide.