Swing Force Calculation: Solve for Horizontal Force

[Mo]

Just a quick question which i don't seem to get the right answer for.

"A parent holds a small child in a swing so that the angle of the swing support is at 20 degrees to the vertical.The child's weight is 200N and the swing seat weighs 50N.What horizontal force must the parent apply"

I have drawn out what it think the vector diagram should look like.

http://img10.exs.cx/img10/1613/diagram6na.gif

I don't think its right.

Assuming that it is .. we would need to use sin/cos/tan. We have both the adjacent and the opposite.So that means we would need to use "Tan"

Tan 70 = 250/adjacent

250/Tan 70 = adjacent (horizontal) which would = 90.99 N .. the answer in the book however is 72.8 N ...

Some help please!

Regards,
Mo

 

[arildno]

What you need to use, is equilibrium of torques about (most naturally) the supporting hinge.

The tension in the string will adapt itself so that equilibrium of forces is established.

So, you must compute the torques of gravity and the horizontally applied force, and ensure that no net torque appears (assume a massless string).

 

[Mo]

Thanks for the reply, i think i can understand what you are saying ... but i still seem to be going round in circles getting the same answer.Is there anyway you could explain a little more please.Even a very quick , rough workthrough would be apreciated.

thanks

Regards,
Mo

 

[arildno]

All right:
Let $$\vec{r}=L(\sin\theta\vec{i}-\cos\theta\vec{j}),\theta=20$$
be the distance vector from the hinge down to the child.
Let $$\vec{W}=-(M_{child}+M_{seat})g\vec{j}$$
be the combined weight, and $$\vec{F}=F\vec{i}$$
be the horizontal force applied.
The equilibrium of moments about the hinge requires:
$$\vec{r}\times\vec{W}+\vec{r}\times\vec{F}=\vec{0}$$
Or:
$$-L\sin\theta(M_{c}+M_{s})g\vec{k}+L\cos\theta{F}\vec{k}=0\vec{k}$$
Or:
$$F=\tan\theta(M_{c}+M_{s})g$$