A block of mass M slides down a plane at an angle of 37 degrees to the horizonta

Question

A block of mass M slides down a plane at an angle of 37 degrees to the horizontal. The coefficient of kinetic friction = 0.20. The horizontal lines in the picture represent magnitude of forces, in units of the force of gravity on the block. The arrows show directions for forces. DATA: sin(37o)=0.60 cos(37o)=0.80, tan(37o)=0.75.

- Enter the two-letter combination for the Normal force on the block. [The two-letters for the force of gravity on the block are LN. Enter L first, (magnitude) and N second, (direction).]

- Enter the 2-letter combination for the frictional force on the block.

Explanation / Answer

a). Normal force |FN| = |JB|*cos(37) = 0.80|JB| = 0.64Mg|B| = 0.64Mg (J = 0.80Mg, |B|=1) 0.64Mg is R so the Normal force in two letters is : RL b) the frction force is Fr = |FN| * 0.20 = 0.20*0.64Mg|B| = 0.128Mg |B| so the friction force in two letters form is: OD O is 0.128Mg, D is the direction.

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