Solve Fnet = ma Problem: Force & Acceleration of Blocks A & B

  • Thread starter A_lilah
  • Start date
In summary: The only thing changing is which one is being pushed with the force. The total mass of the system is always the same, so the acceleration should be the same as well.
  • #1
A_lilah
53
0

Homework Statement


Blocks A and B are sitting side by side (touching) on a horizontal surface. A constant horizontal force Fa is applied to block A, which pushes against block B with a 32.0 N force directed horizontally to the right. The same force Fa is applied to block B; now block A pushes on block B with a 16.0 N force directed horizontally to the left. The blocks have a combined mass of 30.0 kg.

(a) What is the magnitude of the acceleration of the blocks in figure (a)?
(b) What is the magnitude of the force Fa?

Homework Equations



Fnet = ma

The Attempt at a Solution




Part 1: Fa pushes on block A:
Block A has Fa pushing to the right, and block B pushing to the left so

Fnet = massblockA *acceleration = Fa - 32N


Block B pushes back on block A (only horizontal force on it)

Fnet = massblockB *acceleraton = 32N

mass blockA + massblockB = 30kg

Using these three equations I solved for Fa and got that Fa = 30kg*acceleration...
I thought that I could use the next set of data (Fa pushing on block B) to come up with another equation, then I could solve for Fa and acceleration, but the equations are exactly the same, so pretty useless.
Any help would be great, thanks!
(ps~ I sometimes have trouble with FBDs, so if something is wrong with how I laid out the forces, it would be great to know!)
 
Physics news on Phys.org
  • #2
A_lilah said:
I thought that I could use the next set of data (Fa pushing on block B) to come up with another equation, then I could solve for Fa and acceleration, but the equations are exactly the same, so pretty useless.
They aren't exactly the same. Write an equation for the force on block A using the second set of data.
 
  • #3
Ok, Now I'm really not so sure:
For Fa pusing on block A:

massA + massB = 30kg
massblockA *acceleration = Fa - 32N
massblockB *acceleraton = 32N

so,

(Fa - 32)/ acceleration = massA
32/acceleration = mass B

(Fa - 32) + 32 = 30 * acceleration
So Fa = 30 * acceleration

For Fa pusing on B:

massA = 16N / acceleration
masB = (Fa -16) / acceleration
16N + (Fa - 16) = 30 * acceleration
or, Fa = 30 * acceleration

... am I having a math problem then?
 
  • #4
A_lilah said:
Ok, Now I'm really not so sure:
For Fa pusing on block A:

massA + massB = 30kg
massblockA *acceleration = Fa - 32N
massblockB *acceleraton = 32N

so,

(Fa - 32)/ acceleration = massA
32/acceleration = mass B

(Fa - 32) + 32 = 30 * acceleration
So Fa = 30 * acceleration

For Fa pusing on B:

massA = 16N / acceleration
masB = (Fa -16) / acceleration
16N + (Fa - 16) = 30 * acceleration
or, Fa = 30 * acceleration

... am I having a math problem then?
Combine the two equations I highlighted.
 
  • #5
Oh... how did you know to use those equations? (I have a final coming up, and any tips would be great)
Thanks for your help!
 
  • #6
A_lilah said:
Oh... how did you know to use those equations?
To me, those two equations are the most direct mathematical statements of the given data:

A_lilah said:
Blocks A and B are sitting side by side (touching) on a horizontal surface. A constant horizontal force Fa is applied to block A, which pushes against block B with a 32.0 N force directed horizontally to the right.
That tells me that: 32.0 = massB*a

The same force Fa is applied to block B; now block A pushes on block B with a 16.0 N force directed horizontally to the left.
From Newton's 3rd law, I immediately know the force on block A is 16.0 N.
That tells me that: 16.0 = massA*a

Add those equations to get:
32 + 16 = (massA + massB)*a

Then apply:
The blocks have a combined mass of 30.0 kg.
To get the acceleration.

The best thing to do is to write clear FBD diagrams of both blocks for both sets of data. Write all the equations and look for any easy ways to combine them.
 
  • #7
Thanks for your help :)
 
  • #8
Doc Al said:
Combine the two equations I highlighted.

Sorry if I bring it up, but I have the same problem, and I'm trying to figure out how to solve it. I read all the post above, and one thing I don't understand is, the acceleration in those 2 figures should be different, they can't be the same, and if they're not the same, how can we set up the equations ?
 
  • #9
unloved ... did you draw FBD ?

EDIT:first assume that the acceleration is different in two cases, when you set up equations , you will see that they turn out to be same
 
Last edited:
  • #10
theunloved said:
Sorry if I bring it up, but I have the same problem, and I'm trying to figure out how to solve it. I read all the post above, and one thing I don't understand is, the acceleration in those 2 figures should be different, they can't be the same, and if they're not the same, how can we set up the equations ?
Why do you think the accelerations of the two blocks are different? They are in constant contact.
 

What is the formula for Fnet = ma?

The formula for Fnet = ma is the second law of motion, also known as Newton's second law. It states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.

How do I solve for the net force (Fnet)?

To solve for the net force, you will need to know the mass of the object (m) and its acceleration (a). Then, you can simply multiply the mass by the acceleration to find the net force.

How do I find the acceleration (a) in Fnet = ma?

The acceleration (a) in Fnet = ma can be found by dividing the net force (Fnet) by the mass (m). This will give you the value of acceleration in meters per second squared (m/s²).

What units should be used for Fnet and acceleration in Fnet = ma?

The units for Fnet are typically expressed in Newtons (N), which is a unit of force. The units for acceleration are usually given in meters per second squared (m/s²).

Is Fnet always equal to ma in every situation?

Yes, according to Newton's second law, Fnet is always equal to ma. This means that the net force acting on an object is directly proportional to its mass and acceleration. However, it is important to note that this law only applies to objects moving in a straight line at a constant acceleration.

Similar threads

  • Introductory Physics Homework Help
Replies
13
Views
936
  • Introductory Physics Homework Help
Replies
5
Views
1K
  • Introductory Physics Homework Help
Replies
9
Views
12K
  • Introductory Physics Homework Help
Replies
4
Views
885
  • Introductory Physics Homework Help
Replies
4
Views
3K
  • Introductory Physics Homework Help
Replies
2
Views
666
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
11
Views
1K
  • Introductory Physics Homework Help
Replies
7
Views
329
  • Introductory Physics Homework Help
Replies
23
Views
2K
Back
Top