Problem with Newton's second law

[Bohrok]

Homework Statement

In the following figure, the boy has a mass of 40kg, and the platform he is sitting on has a mass of 15kg. If the force of the board on the boy is 180N;

a) Find the acceleration of the boy

b) Find the reading of the scale.

http://www.csupomona.edu/~jarmand/131/131chall5s12_files/image002.gif

Homework Equations

F = ma

The Attempt at a Solution

I think I have the acceleration kind of worked out with ƩF_y = 2T - mg = ma_y where m is the total mass of the system (platform + boy) and T is the tension in the rope. Would the force of the tension be the same as the reading on the scale?

Mostly I've been unsure how to use that 180N. Then I started thinking that the downward force that the boy exerts on the chair has the same magnitude as the upward normal force that the chair exerts on the boy, giving ƩF_y = T + F_n - m_boyg = m_boya_y
Two equations with two unknowns T and a:
2T -55(9.8) = 55a
T + 180 - 40(9.8) = 40a
?

 

[haruspex]

That all looks correct. Do you not know how to solve simultaneous equations? Just get one of the equations in the form T = (some function of other variables) and use this to substitute for T in the other equation.

 

[Bohrok]

Oh I know how to solve simultaneous equations; I just wanted to see if my work seemed alright
Thanks!

If no one else chimes in, I'll take it that they agree

 

[SammyS]

Homework Statement

In the following figure, the boy has a mass of 40kg, and the platform he is sitting on has a mass of 15kg. If the force of the board on the boy is 180N;

a) Find the acceleration of the boy

b) Find the reading of the scale.

[ IMG]http://www.csupomona.edu/~jarmand/131/131chall5s12_files/image002.gif

Homework Equations

F = ma

The Attempt at a Solution

I think I have the acceleration kind of worked out with ƩF_y = 2T - mg = ma_y where m is the total mass of the system (platform + boy) and T is the tension in the rope. Would the force of the tension be the same as the reading on the scale?

Mostly I've been unsure how to use that 180N. Then I started thinking that the downward force that the boy exerts on the chair has the same magnitude as the upward normal force that the chair exerts on the boy, giving ƩF_y = T + F_n - m_boyg = m_boya_y
Two equations with two unknowns T and a:
2T -55(9.8) = 55a
T + 180 - 40(9.8) = 40a
?

Yes. That all looks fine to me. I got the same thing without first looking at your solution.

 

[Nickie]

First of all, let me clarify that Newton's second law states that the net force acting on an object is equal to its mass multiplied by its acceleration. So in this problem, we are looking for the acceleration of the boy.

To solve this, we can use the equations you have mentioned, ƩFy = 2T - mg = may and ƩFy = T + Fn - mboyg = mboyay. However, there are a few things to consider.

Firstly, the force of the board on the boy (180N) is not the only force acting on the boy. There is also the force of gravity (mboyg) and the normal force (Fn) exerted by the chair on the boy. So our equation for ƩFy would be 2T - mboyg - Fn = mboyay.

Secondly, the tension in the rope (T) is not the same as the reading on the scale. The tension in the rope is the force that the rope exerts on the platform, while the reading on the scale is the normal force that the platform exerts on the boy. These two forces are equal in magnitude but opposite in direction.

So, to solve for the acceleration of the boy, we can use the following steps:

1. Draw a free-body diagram for the boy, including all the forces acting on him.

2. Write out the equation for ƩFy, taking into account all the forces.

3. Use the given values (mboy = 40kg, mplatform = 15kg, Fboard = 180N) to solve for the unknowns (T and a).

4. To find the reading on the scale, we can use the equation Fn = mboyg + mboyay. Plug in the values we have found for a and solve for Fn.

I hope this helps clarify any confusion you may have had with using Newton's second law in this problem. Remember, it is always important to draw a clear free-body diagram and consider all the forces acting on an object when solving problems like this. Keep up the good work!