When 10 N force applied at 30 degrees to the end of a 20 cm handle of a wrench, it was just able to loosen the nut. What magnitude of the force would require to just loosen the nut, of the force apply perpendicularly at the end of the handle

5 N

Explanation:

From the question,

The magnitude of the force that would be required to just loosen the nut when the force is applied perpendicularly at the end of the handle is

Fy = Fsinθ Equation 1

Where Fy = force acting perpendicular at the end of the handle, F = Force applied to the handle, θ = angle of inclination of the force to the end of the handle.

Given: F = 10 N, θ = 30°

Substitute these values into equation 1

Fy = 10(sn30°)

Fy = 10(0.5)

Fy = 5 N.

it is a negative acceleration