What is the Force Exerted by a Child on a Swing Chair?

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In summary, the force of a child on a swing can be calculated using Newton's Second Law of Motion and varies based on the child's weight and the speed of the swing. This force is not constant and can change due to changes in weight, speed, and external factors. The force affects the swing by causing it to move and can be measured using tools such as a force gauge or estimated by observing the height and speed of the swing.
  • #1
myelevatorbeat
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Homework Statement


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.

Homework Equations


F=ma


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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  • #2
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
 
  • #3
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.
 
  • #4
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.
 
  • #5
myelevatorbeat said:
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
 
Last edited:
  • #6
Always draw a freebody diagram of the body being analyzed. In this case, the chair.
 
  • #7
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??
 
  • #8
myelevatorbeat said:
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. :wink: Sorry, I should have pointed it out.
 
Last edited:
  • #9
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
 
  • #10
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.
 
  • #11
myelevatorbeat said:
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.
 
  • #12
rootX said:
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.
 
Last edited:
  • #13
I got it, it was supposed to be carried out to 111.63 but I had rounded to 112.
 
  • #14
myelevatorbeat said:
I got it, it was supposed to be carried out to 111.63 but I had rounded to 112.

Cool. :smile:
 

1. What is the force of a child on a swing?

The force of a child on a swing can be calculated using Newton's Second Law of Motion, which states that force (F) equals mass (m) multiplied by acceleration (a). In this case, the mass refers to the weight of the child and the acceleration refers to the change in speed or direction of the swing. The force of the child on the swing will vary depending on the child's weight and the speed at which they are swinging.

2. How does the force of a child on a swing change?

The force of a child on a swing can change in two ways: through changes in the child's weight and changes in the speed of the swing. If the child's weight increases, the force on the swing will also increase. Similarly, if the speed of the swing increases, the force on the swing will also increase. This is because both weight and speed are factors in determining the force according to Newton's Second Law.

3. Is the force of a child on a swing constant?

No, the force of a child on a swing is not constant. As the child swings back and forth, the force will change due to changes in weight and speed, as mentioned in the previous question. Additionally, the force may also change depending on external factors such as wind resistance or friction between the swing and the air.

4. How does the force of a child on a swing affect the swing itself?

The force of a child on a swing affects the swing by causing it to move. As the child pushes and pulls on the swing, they are exerting a force on it. This force causes the swing to accelerate in the direction of the force. The greater the force, the greater the acceleration and therefore, the higher and faster the swing will move.

5. Can the force of a child on a swing be measured?

Yes, the force of a child on a swing can be measured using a variety of tools such as a force gauge or a scale. These tools can measure the weight of the child and the speed of the swing, which can then be used to calculate the force. Alternatively, the force can also be estimated by observing the height and speed of the swing and making comparisons to known forces.

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