Its all about forces... show please the solution

Question

a 5kg block rests on a 30deg. incline. the coefficient of static friction between the block and the incline is 0.20. How large a horizontal force must push on the block if the block is to be on the verge of sliding up the incline?

Answer 1

In free body diagram you need a force of {5X9.8Sin 30}+{5X9.8Cos 30}X

0.20 to just start moving.So you need (24.5) +(42.435)X0.20=24.5 +8.487 = 32.987 newtons.Now your hyptonus is 32.987 newtons,you need more force to push it horizontally so 32.987/Cos 30=38.1 newtons.

Answer 2

From the free body diagram of the problem, the following working equations can be set up:

Equation 1 : summation of forces along the x-axis (axis parallel to the incline)

F - f - mgsin 30 = 0

where

F = force required to push the block up the incline

f = frictional force

m = mass of the block = 5 kg (given)

g = acceleration due to gravity = 9.8 m/sec^2 (constant)

Equation 2 : summation of forces along the y-axis (axis perpendicular to the incline)

N - mgcos 30 = 0

N = mgcos 30

where

N =  normal force exerted by the plane on the block

and all the other terms have been previously defined.

Equation 3 -- by definition,

f = frictional force = coeff of friction * N

and since N = mgcos 30 (from Equation 2)

f = 0.20(5)(9.8)(cos 30)

f = 8.49 N

Substituing f = 8.49 in Equation 1,

F - 8.49 - (5)(9.8)(sin 30) = 0

and solving for F,

F = 8.49 + 5(9.8)(sin 30)

F = 32.99 N