Tensions in a swing problems using variables

[Confusingmeh]

Homework Statement

An adult exerts a horizontal force on a swing that is suspended by a rope of length L, holding it at an angle \Theta with the vertical. The child in the swing has a weight W and dimensions that are negligible compared to L. The wights of the rope and of the seat are negligible.
In terms of W and \Theta , determine:

a) the tension in the rope

b) the horizontal force exerted by the adult

The adult releases the swing from rest. In terms of W and \Theta determine:

c) the tension in the rope just after the release (the swing is instantaneously at rest)

d) the tension in the rope as the swing passes through its lowest point

Homework Equations

The Attempt at a Solution

ok so I am trying to check if I'm right for these problems...

a) T = W + W(tan\Theta)

b) F = W(tan\Theta)

c) T = W + W(tan\Theta)

d) less sure about this one teacher emailed and said we didnt have do it because we havnt learned this yet but i got:
T = W + m(19.8{tan\Theta}{(\Theta 2 \pi)/(360)})

 

[collinsmark]

A few things to remember,

Draw your free body diagram (FBD). Parts a) and b) can share the same FBD for this problem since all the forces are identical, but you'll need to draw a new one for part c).

Remember Newton's second law of motion.

• For a given direction (x- or y-component), ma = sum of all forces in that direction (for the acceleration component a in the same direction).
• For parts a) and b), nothing is accelerating (static equilibrium), so the sum of all forces [in any given direction] equals zero.
• For part c) things are accelerating, so the sum of all forces do not add up to zero in both directions. There is a "resultant" net force.

Also, when looking at your FBD, always remember the following:

\sin \theta = \frac{\mathrm{opposite}}{\mathrm{hypotenuse}}

\cos \theta = \frac{\mathrm{adjacent}}{\mathrm{hypotenuse}}

\tan \theta = \frac{\mathrm{opposite}}{\mathrm{adjacent}} = \frac{\sin \theta}{\cos \theta}

ok so I am trying to check if I'm right for these problems...

a) T = W + W(tan\Theta)

You'll need to redo part a). Take a look at your FBD again.

b) F = W(tan\Theta)

Part b) looks good to me!

c) T = W + W(tan\Theta)

Remember, you need to draw a new FBD, because the adult's horizontal force is no longer there. You'll have to redo part c)

d) less sure about this one teacher emailed and said we didnt have do it because we havnt learned this yet but i got:
T = W + m(19.8{tan\Theta}{(\Theta 2 \pi)/(360)})

I think you might be on the right idea, generally speaking, for part d), but I got something quite a bit different. You seem to be converting units (radians to degrees or some-such). But that's not necessary. Θ is Θ, whatever units that might be. And I think you're using the wrong trigonometric function. But if your instructor says that part d) is not necessary, perhaps we can skip part d) for now.