DUE TODAY ! A block of mass 4kg is pushed up against a wall by a force (P) that makes the 50 degree angle with the horizontal.

A. Calculate the force P needed if the coefficient of static friction between the block and the wall is 0.

B. Calculate the the force or forces needed if the coefficient 0.25.

Please answer in Newton’s with two 2 significant figures.

I suppose you mean to say that the wall is frictionless in the first scenario? Also, I assume the block is to held in place. Construct a free body diagram for the block. There are 3 (in part A) or 4 (in part B) forces acting on it.

• its weight w, pulling it downward

• the normal force (magnitude n), pushing outward from the wall to the left

• the push as described, with magnitude p

• static friction (mag. f ), opposing the upward net force and thus pointing downward.

The static friction force has a magnitude proportional to that of the normal force. If the coefficient of static friction is µ, then

f = µ n

so if the wall is frictionless with µ = 0, then f = 0 and does not need to be considered.

(A) If µ = 0, then by Newton's second law we have

• net vertical force:

∑ F = p sin(50°) - w = 0

and we don't need to consider the net horizontal force to determine p from here. We get

p = w / sin(50°) = (4 kg) (9.8 m/s²) / sin(50°) ≈ 51 N

(B) If µ = 0.25, then Newton's second law gives

• net vertical force:

∑ F = p sin(50°) - w - f = 0

p sin(50°) - f = (4 kg) (9.8 m/s²)

p sin(50°) - f = 39.2 N

• net horizontal force:

∑ F = p cos(50°) - n = 0

p cos(50°) - f /0.25 = 0

[since f = 0.25 n]

p cos(50°) - 4f = 0

Multiply the first equation by -4, then add it to the second equation to eliminate f and solve for p :

-4(p sin(50°) - f ) + (p cos(50°) - 4f ) = -4 (39.2 N) + 0

p (cos(50°) - 4 sin(50°)) = -156.8 N

p = (156.8 N) / (4 sin(50°) - cos(50°)) ≈ 65 N

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hope this !