Question

A horizontal force, F1 = 95 N, and a force, F2 = 18.9 N acting at an angle of θ to the horizontal, are applied to a block of mass m = 2.9 kg. The coefficient of kinetic friction between the block and the surface is μk = 0.2. The block is moving to the right. Calculate the normal force when the angle is 30. and acceleration when the angle is 30.

Answer 1

Newton's second law:

ΣF = m x a

Where:

ΣF = sum of forces

m = mass = 2.9 kg

a = acceleration

x-axis = parallel to the movement of the block.

y-axis = perpendicular to x axis

m= 2.9 kg

F1 (horizontal ) = 95N

F2 (at 30°) = 18.9

μk = 0.2

g=9.8 m/s^2 (gravity)

weight = m * g = 2.9 kg* 9.8 m/s^2 _=28.42 N

Calculate the componets of F2

F2x = F2*cos 30 = 18.8 *cos 30 = 16.37N

F2Y = 18.8* sin 30 = 9.4 N

Normal force:

ΣFy = m x ay

aY= 0

FN - F2y - W = 0

FN + 9.4 N - 28.42 N = 0

FN = +9.4 + 28.42

FN = 37.82

Friction force = μk * FN = 0.2 * 37.82= 7.56N

Sum of horizontal forces:

ΣFx = 95N +16.37-7.56 = 2.9 * a

F = m*a

a = F/m

103.81N = 2.9 *a

103.81 /2.9 = a

a= 37.79 m/s^2