# Thread: need help with a force diagram question

1. ## need help with a force diagram question

This is probably a very simple question, but I just don't know where to start with it.

I need to show that and are related by the following equation:

by creating a force diagram of the situation shown in the attachment.

Any help at all would be greatly appreciated

P.S. I have also posted on Physics Help Forum as I am not quite sure where this particular question fits in.

2. Originally Posted by trent19
This is probably a very simple question, but I just don't know where to start with it.

I need to show that and are related by the following equation:

by creating a force diagram of the situation shown in the attachment.

Any help at all would be greatly appreciated

P.S. I have also posted on Physics Help Forum as I am not quite sure where this particular question fits in.
The posted diagram is misleading. I thought that the straight members of the system are more massive than the mass m at the end of the pivoted h member.
But the equation that you want to prove implies that the masses of the straight members be ignored. So let us consider the pivoted h member as a sling only.
Let us imagine the mass m as a bob whose mass is m.

The rotational speed of the axle induces centripetal forces to the members of the system proportional to the distance of the members from the axle, and those inward forces are perpendicular to the axle. Since the axle is vertical, then the centripetal forces are horizontal.

On the bob, the centripetal force is
Fx = m*(v^2)/r ------------(i)
where
v = tangential velocity of the bob
r = horizontal distance of the bob from the axle

v = w*r
So,
(v^2)/r = (w*r)^2 / r = (w^2)*r

r = a +h*sin(theta)

Hence,
Fx = m[(w^2)*(a +h*sin(theta)] --------(ii)

On the bob also, the gravity exerts a vertical force
Fy = weight = m*g -------(iii)

Now cut the h string at the pivot. Supply necessary forces to the string at that cut point to keep the severed members in equilibrium. Meaning, draw a free-body diagram of the severed portion of the system.
To ignore whatever are those supplied forces at the cut point, we take moments at that cut point.

The FBD, (free-body diagram), shows two forces acting on the bob:
>>>Fx, which is counteracting the centripetal force Fx, is pointing horizontally away from the axle.
>>>Fy, the vertical force pointing vertically downward because of the weight of the bob.

Summation of moments at the cut point, at the pivot, is zero.
(Fx)*(h*cos(theta)) -(Fy)*(h*sin(theta)) = 0
Substitutions,
[m(w^2)(a +h*sin(theta)](h*cos(theta) -[mg](h*sin(theta)) = 0
(mh)(w^2)cos(theta)[a +h*sin(theta)] = (mh)(g)sin(theta)
The (mh) cancels out, then divide both sides by cos(theta)
(w^2)[a +h*sin(theta)] = (g)sin(theta) / cos(theta)
[a +h*sin(theta)](w^2) = (g)tan(theta) --------------proven.