Physics: Normal Forces Problem?

Question

An ant walks away from the top of a ball. If the ant starts to slip when the normal force on its feet drops below one-half its weight, at what angle does slipping begin?

Note: No frictional forces
I understand that N=1/2m in the equation N=mg, therefore 1/2m=mg and 1/2=g , but I do not know how to solve from there.

Answer 1

> I understand that N=1/2m ...

In that case you're missing something. You have set a force (N) equal to a mass (1/2 m). This means there's a mistake in your units; and that's leading to some further mistakes.

> ...in the equation N=mg ...

The equation "N=mg" is true only when something is resting on a HORIZONTAL surface (like the very TOP of the ball). When the surface is inclined at an angle θ from the horizontal, the formula is: "N = mg(cosθ)"

Here's how I would approach it.

Ant's Weight:
W = mg

Normal force at slip:
N = mg(cosθ)

According to the problem,
normal force at slip = 1/2 weight
N = (1/2)W
mg(cosθ) = (1/2)mg

or:
cosθ = 1/2

Now just solve for θ

Answer 2

Draw a line from the center of the ball to the top of the ball. This vertical line is the normal at the top point. Rotate this radius vector through an angle θ about the axis through the center of the circle to a new point P.

The normal N at P is inclined at angle θ from the vertical line at the center of the circle.

The weight mg acts vertically down at this new point

The component of mg along the normal = mg sin θ = 0.5 mg (given)

This Implies θ = 30 degree.

Answer 3

well the normal force is always perpendicular to the surface. so as the angle increases the normal force becomes less vertical and is split into vertical and horizontal components. I assume you know how to solve for those using trig. so at what angle is the vertical component less than half of the ant's weight?