Force of child on a swing


by myelevatorbeat
Tags: child, force, swing
myelevatorbeat
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#1
Sep8-07, 07:14 PM
P: 55
1. The problem statement, all variables and given/known data
An inventive child wants to reach an apple in a tree without climbing the tree. Sitting in a chair connected to a rope that passes over a frictionless pulley (Fig. P4.79), the child pulls on the loose end of the rope with such a force that the spring scale reads 350 N. The child's true weight is 310 N, and the chair weighs 160 N.

(a) Show that the acceleration of the child-chair system is upward and find its magnitude.

(b) Find the force the child exerts on the seat of the chair.

2. Relevant equations
F=ma


3. The attempt at a solution

I solved for Part A and got it correct by going:

700-470=47.96a
a=230/47.96
a=4.7956 which I rounded to 4.8 m/s^2


I'm having trouble with Part B. Here is the work I did so far:

310-350=-40
-40+160=120 N
350-120 N = 230 N

However, this is wrong and I'm not sure where I went wrong. If someone could help me, I'd be very thankful.
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learningphysics
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#2
Sep8-07, 07:34 PM
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P: 4,125
For part b, what are the forces acting on the chair? You should have 3 forces. Use net force = ma, for the chair alone... you know a = 4.8m/s^2
Astronuc
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#3
Sep8-07, 07:34 PM
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I assume the spring scale is connected to the rope?

B. The chair is being accelerated under the child, so the mass of the child experiences the acceleration of gravity and the acceleration of the chair. So the weight of the child must include both accelerations.

Meanwhile the child is pulling on the rope, with a force of 350 N, which you did correctly.

myelevatorbeat
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#4
Sep8-07, 07:46 PM
P: 55

Force of child on a swing


I assume the three forces are 700 N (tension in string), 310 N (child's weight), and 160 N (weight of the chair).

So, I set it up like this

F=ma
700 N - 160 N - 310 N = (160/9.8) x 4.80
which gives me
230 N = 78.384

I'm really confused about what I should do at this point.
learningphysics
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#5
Sep8-07, 07:55 PM
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Quote Quote by myelevatorbeat View Post
I assume the three forces are 700 N (tension in string), 310 N (child's weight), and 160 N (weight of the chair).

So, I set it up like this

F=ma
700 N - 160 N - 310 N = (160/9.8) x 4.80
which gives me
230 N = 78.384

I'm really confused about what I should do at this point.
The 310N force isn't right.

The trick to these force problems is not to assume anything. Go with only what you know for sure. What are the forces exerted on the chair? The rope is connected to the chair. It exerts a 350N force on the chair (I'm assuming one end is connected to the chair, and the girl holds the other end right?). Gravity exerts a force of 160N on the chair. And the girl exerts a force whose magnitude you need to find out. Don't assume it's the weight of the girl... you don't know that.

so
F = ma
350N - 160N - Fgirl = ma

solve for Fgirl
learningphysics
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#6
Sep8-07, 07:59 PM
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P: 4,125
Always draw a freebody diagram of the body being analyzed. In this case, the chair.
myelevatorbeat
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#7
Sep8-07, 08:05 PM
P: 55
OK

I set it up like this:

700 N - 160 N - Fgirl=(160/9.8) x .4.80

540 N = 78.367 + Fgirl
Fgirl=462 N

This is still wrong though, so maybe I should be using 350 as my tension force instead of the 700 I used in the previous part of the question??
learningphysics
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#8
Sep8-07, 08:06 PM
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Quote Quote by myelevatorbeat View Post
OK

I set it up like this:

700 N - 160 N - Fgirl=(160/9.8) x .4.80

540 N = 78.367 + Fgirl
Fgirl=462 N

This is still wrong though, so maybe I should be using 350 as my tension force instead of the 700 I used in the previous part of the question??
Yup. See the equation in my post above. Sorry, I should have pointed it out.
rootX
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#9
Sep8-07, 08:08 PM
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Hey, for part a
can anyone explain why it's 2(350)-W=Net force?

I draw FBD, and got like this

Fchild - T = Fnet
T-W = Fnet
myelevatorbeat
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#10
Sep8-07, 08:11 PM
P: 55
Ok, so now I set it up this way:
350 N - 160 N -Fgirl=(160/9.8) x 4.80 m/s^s
and when I solved for Fgirl I got 111.6327

Webassign says this is wrong still so I must still be doing something wrong.
learningphysics
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#11
Sep8-07, 08:17 PM
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Quote Quote by myelevatorbeat View Post
Ok, so now I set it up this way:
350 N - 160 N -Fgirl=(160/9.8) x 4.80 m/s^s
and when I solved for Fgirl I got 111.6327

Webassign says this is wrong still so I must still be doing something wrong.
I'm getting that also. Maybe since it's acting downward it should be -111.6327N? Not sure... hope someone else clarifies.
learningphysics
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#12
Sep8-07, 08:18 PM
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Quote Quote by rootX View Post
Hey, for part a
can anyone explain why it's 2(350)-W=Net force?

I draw FBD, and got like this

Fchild - T = Fnet
T-W = Fnet
For that part, you consider the pulley and child together as one system, and take the freebody diagram of that. You have the two ends of the rope both with a tension of 350N.
myelevatorbeat
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#13
Sep8-07, 08:21 PM
P: 55
I got it, it was supposed to be carried out to 111.63 but I had rounded to 112.
learningphysics
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#14
Sep8-07, 08:22 PM
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Quote Quote by myelevatorbeat View Post
I got it, it was supposed to be carried out to 111.63 but I had rounded to 112.
Cool.


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