Solving Part 2 of Homework: Mass, Force, and Acceleration

[rodjav305]

Homework Statement

Homework Equations

a=F/m

The Attempt at a Solution

For part 1 I did this :

Mass of boy: 353 N/9.8m/s^2 = 36.020408 kg
Mass of chair: 64 N/9.8m/s^2 = 6.530612 kg

Then :

(36.020408 + 6.530612)a = 2(213.5)-353-64= 10N

a= 10 N/ (36.020408 + 6.530612) = 0.235012 m/s^2

I need help on part 2, any help is appreciated!

 

[PhanthomJay]

Homework Statement

The Attempt at a Solution

For part 1 I did this :

Mass of boy: 353 N/9.8m/s^2 = 36.020408 kg
Mass of chair: 64 N/9.8m/s^2 = 6.530612 kg

Then :

(36.020408 + 6.530612)a = 2(213.5)-353-64= 10N

a= 10 N/ (36.020408 + 6.530612) = 0.235012 m/s^2

very good!

I need help on part 2, any help is appreciated!

Now you need to draw a free body diagram of the chair, in order to calculate the normal force of the boy on the chair, OR, draw a free body diagram of the boy, in order to determine the normal force of the chair on the boy. You can do either FBD, whichever one you find easier to identify the forces acting, then use Newton's laws. Note that the acceleration of the boy and the acceleration of the chair are the same as the acceleration of the system.

 

[rodjav305]

Ok so i drew a FBD, i got the force of the boy minus the force of the chair (36.020408 -6.530612) = 29.489796 N exerted on the chair...Is that right?

 

[rodjav305]

So that result is wrong.
Another idea that i have is take all the forces and subtract it and get that as a mass and multiply with acceleration...

36.020408-21.785714-6.530612 = (7.704082)(0.235012)= 1.810552 N

I don't know if that wrong, it looks wrong to me because of the little force exerted on the chair... help please :)

 

[PhanthomJay]

Ok so i drew a FBD, i got the force of the boy minus the force of the chair (36.020408 -6.530612) = 29.489796 N exerted on the chair...Is that right?

I am not sure how you answered part a so nicely, while for this part b you are for some reason subtracting masses in units of kg and ending up with a force in units of Newtons, and then ignoring Newton's laws as well

So that result is wrong.
Another idea that i have is take all the forces and subtract it and get that as a mass and multiply with acceleration...

36.020408-21.785714-6.530612 = (7.704082)(0.235012)= 1.810552 N

I don't know if that wrong, it looks wrong to me because of the little force exerted on the chair... help please :)

Yes, it is wrong, I don't know what you are doing here.
It is a bit difficult to draw a free body diagram of the boy or the chair in this example, but nontheless, let's focus on the forces acting on the boy. His weight acts down, there is a tension force acting up on him, and the contact normal force of the chair is pushing up on him. The algebraic sum of these forces is the net force acting on him. Now use Newton's 2nd law, F_net =ma, where m is the mass of the _____?

Alternatively, you can look at the forces ating on the chair. Its weight acts down, there is a tension force acting up on it, and the contact normal force of the boy is pushing down on it. The algebraic sum of these forces is the net force acting on the chair. Now use Newton's 2nd law, F_net =ma, where m is the mass of the _____?

 

[rodjav305]

Ok so m would be the mass of the boy, chair and spring(addition of all of the masses)?

Or just adding the mass of the boy and subtracting the tension and chair mass. For acceleration i would use the result i got from a and plug it in.

If i add up all the masses:
36.020408 + 6.530612 + 21.785714 = 64.336734 kg * 0.235012 m/s^2 = 15.119905 N

The result would be the force exerted on the chair..this result seems right to me, but I'm not sure?

 

[PhanthomJay]

Ok so m would be the mass of the boy, chair and spring(addition of all of the masses)?

. No.

First, the spring has negligible mass and therefore negligible weight. It just serves here to mesure the tension force in the rope.

Or just adding the mass of the boy and subtracting the tension and chair mass. For acceleration i would use the result i got from a and plug it in.

If i add up all the masses:
36.020408 + 6.530612 + 21.785714 = 64.336734 kg * 0.235012 m/s^2 = 15.119905 N

The result would be the force exerted on the chair..this result seems right to me, but I'm not sure?

No, incorrect. I have absolutely no idea how you solved part a unless it was given you. But as for part b, when you isolate a mass, like the chair, you examine all the forces acting on it, and use that mass alone in your net force =ma equation. I've already given you some hints in post #5.

 

[rodjav305]

So all i would need to do it add up the masses of the boy and the chair which is 42.55102 kg and multiply by acceleration which should give me the result 10.0 N

 

[PhanthomJay]

So all i would need to do it add up the masses of the boy and the chair which is 42.55102 kg and multiply by acceleration which should give me the result 10.0 N

No, that's the net force acting on the boy and chair that you correctly used in part a.

We're onto part b now.

Lets focus on the forces acting on the boy, Chris. His weight (given) is pulling down on him, there is a tension force (given) pulling up on him, and there is a contact unknown normal force (N) of the chair pushing up on him. The algebraic sum of these forces (take up as positive and down as negative) is the net force acting up on him. Now use Newton's 2nd law, F_net =ma, where m is the mass of Chris, to solve for N, the normal force.

 

[I Like Pi]

Hey, I am doing a homework question just like this one, but i was wondering why you would need to multiply the tension by two? I know for example, the classical two horses pulling on a rope with equal force, so if both horses pull with 100 N, the tension is 100 N not 200 N, but why do we have to multiply the tension by 2 in this case?

thanks for your time

 

[PhanthomJay]

Hey, I am doing a homework question just like this one, but i was wondering why you would need to multiply the tension by two? I know for example, the classical two horses pulling on a rope with equal force, so if both horses pull with 100 N, the tension is 100 N not 200 N, but why do we have to multiply the tension by 2 in this case?

thanks for your time

There's only one rope in the horse pull problem. In this problem, there are two rope forces acting on the child-swing system when you draw a FBD (Free Body Diagram) that isolates the child and swing from the pulley. This is not the case when you draw the FBD showing the forces on the child alone, since only one rope acts on him (at his hands).

 

[I Like Pi]

Ohh, okay, but how do you incorporate the two equations? I got that FT-Fgchair = mchair*a, and FT-Fgchild=mchild*a. I tried to solve for a for both, and then add the two, but i don't get the right answer :/

 

[PhanthomJay]

Ohh, okay, but how do you incorporate the two equations? I got that FT-Fgchair = mchair*a, and FT-Fgchild=mchild*a. I tried to solve for a for both, and then add the two, but i don't get the right answer :/

If you are looking at the forces on the chair, then FT -Fgchair - N =mchair*a. You forgot the normal force of the boy acting down on the chair. If you are looking at the forces on the boy, then FT -Fgchild + N = mchild*a. Again, you forgot the normal force of the chair acting up on the child. FT is given as 213.5 N.

As a check, you can look at the child-chair system, where the tension force 2T acts up, and the weight of the boy and chair act down, solve for a using Newton 2 and using the mass of both the boy and chair (no normal force in this FBD since it is internal to your FBD of the system).

 

[I Like Pi]

If you are looking at the forces on the chair, then FT -Fgchair - N =mchair*a. You forgot the normal force of the boy acting down on the chair. If you are looking at the forces on the boy, then FT -Fgchild + N = mchild*a. Again, you forgot the normal force of the chair acting up on the child. FT is given as 213.5 N.

As a check, you can look at the child-chair system, where the tension force 2T acts up, and the weight of the boy and chair act down, solve for a using Newton 2 and using the mass of both the boy and chair (no normal force in this FBD since it is internal to your FBD of the system).

Oh, thank you that makes sense, but what i don't get is that, i know for the boy himself, he has a normal force because he lies on the swing, which is the swing acting on the boy. What i don't get is why does the chair have a normal force? As it has nothing underneath it? Wouldn't the force of the boy be, the force of the boy? not the normal force? cause then its a reaction force

Thanks for your time!

 

[PhanthomJay]

Oh, thank you that makes sense, but what i don't get is that, i know for the boy himself, he has a normal force because he lies on the swing, which is the swing acting on the boy. What i don't get is why does the chair have a normal force?

By Newton's 3rd law, if the chair exerts a normal force on the boy, the boy must exert an equal and opposite normal force on the chair.

As it has nothing underneath it?

But its got the kid on top of it! Normal forces act perpendicular to the surfaces in contact, and always push toward the object on which they act, if not physicaly connected to it.

Wouldn't the force of the boy be, the force of the boy? not the normal force?

The force of the boy is the normal force on the chair, acting down on it. The force of the boy on the chair is not his weight.

cause then its a reaction force

Thanks for your time!

I guess it's an action/reaction force: the boy exerts an action force (normal force) on the chair, and the chair exerts an equal and opposite reaction force (normal force) on the boy.