A 12-kg crate rests on a horizontal surface and a boy pulls on itwith a force that is 30? below the horizontal. If the coefficient of static friction is 0.40, theminimum magnitude force he needs to start the crate moving is: A. 44N B. 47N C. 54N D. 56N E. 71N

E. 71 N

Explanation:

Given Data

Mass=12 kg

Angle=30°

static friction=0.44 N

To Find

Force=?

Solution

Sum of the forces in the direction of motion is given as

= 0 = F*cos (30) - friction

Sum of the forces in the vertical direction (up is positive) is given as

= 0 = N - mg - F*sin (30)

N = m*g + F*sin (30)

From the definition of friction

friction = u*N

friction = u * (m*g+F*sin (30))

Plug back into the top

0 = F*cos (30) - (u*m*g+u*F*sin (30))

Solve for F

u*m*g = F * (cos (30) - u*sin (30))

F = (u*m*g) / (cos (30) - u*sin (30))

Put the number we get

F = 71 N

So Option E is correct

C. 54N

Explanation:

From Friction

F' = μR ... Equation 1

Where F' = Friction force, μ = Coefficient of static friction, R = Normal reaction

But,

R = mg ... Equation 2

Where m = mass of the crate, g = acceleration due to gravity of the crate.

substitute equation 2 into equation 1

F' = μmg ... Equation 3

Given: m = 12 kg, μ = 0.4, g = 9.8 m/s²

Substitute into equation 3

F' = 12 (0.4) (9.8)

F' = 47.04 N.

From the question, the boy pulls with a force of 30° below the horizontal.

F' = Fcos30°

Where F = The force needed to start the crate moving

make F the subject of the equation

F = F'/cos30

F = 47.04/cos30

F = 47.04/0.8660

F = 54.32 N.

Hence the right option is C. 54N