Three identical blocks connected by ideal strings, Find Horizontal F

[khem student]

Homework Statement

Three identical blocks connected by ideal strings are being pulled along a horizontal frictionless surface by a horizontal force. The masses of the three blocks are 15.1kg , 19.8kg , and 36.0kg , Answer the following questions concerning the motion of the blocks.
1. If the blocks move across the horizontal frictionless surface with a constant velocity of 2.5 m/s, what is the horizontal force?

What I don't understand is that the blocks are listed as Identical but all vary in weight.

Homework Equations

Image is attached.

The Attempt at a Solution

I know Force = mass * gravity otherwise known as 9.81 m/s^2
and I know that if I divide the velocity 2.5 m/s by time it will = mass
I know time= distance/velocity, but no distance is given, so what formula do I use to solve for horizontal force in this situation?
Typically A=B,B=C,C=A if they are all identical blocks, but here the weights vary among the blocks so how would that be taken into account.
The prof tends not to explain the process and just assigns the work...

 

[Doc Al]

What I don't understand is that the blocks are listed as Identical but all vary in weight.

That is certainly odd. Just imagine that they are the same size and shape, but have different masses.

I know Force = mass * gravity otherwise known as 9.81 m/s^2

That gives the weight of a given mass. That may or may not be relevant to this problem.

and I know that if I divide the velocity 2.5 m/s by time it will = mass

Not sure why you think that.

I know time= distance/velocity, but no distance is given, so what formula do I use to solve for horizontal force in this situation?
Typically A=B,B=C,C=A if they are all identical blocks, but here the weights vary among the blocks so how would that be taken into account.

Hint: This is something of a trick question. What net force is required for an object to have a constant velocity?

 

[Doc Al]

1. If the blocks move across the horizontal frictionless surface with a constant velocity of 2.5 m/s, what is the horizontal force?

Are you sure that the problem doesn't state constant acceleration (in m/s^2) instead of constant velocity?

 

[khem student]

Are you sure that the problem doesn't state constant acceleration (in m/s^2) instead of constant velocity?

yes, I am sure it says velocity.

Hint: This is something of a trick question. What net force is required for an object to have a constant velocity?

There is no net force when velocity is constant due to Newton's law, an object in motion continues in that motion unless acted on by an external force.
So does that mean the horizontal force would be = 0 N, because the velocity is constant?

 

[Doc Al]

There is no net force when velocity is constant due to Newton's law, an object in motion continues in that motion unless acted on by an external force.
So does that mean the horizontal force would be = 0 N, because the velocity is constant?

Exactly.

 

[khem student]

It is 0 N! Thank You! I wasn't sure.
Now if an acceleration of 1.5 m/s^2 is applied, how do you find force?
would a variation of the equation a= (Net Force/mass) be valid for this problem?
and if so would i add together the total weight of all blocks and solve or attempt to do them all individually then add the sums together?

 

[Doc Al]

It is 0 N! Thank You! I wasn't sure.

Good.

Now if an acceleration of 1.5 m/s^2 is applied, how do you find force?

By using Newton's 2nd law.

would a variation of the equation a= (Net Force/mass) be valid for this problem?

Sure. That's another statement of ƩF = ma.

and if so would i add together the total weight of all blocks and solve or attempt to do them all individually then add the sums together?

Careful! You're not adding weights, but perhaps the masses.

You can apply Newton's 2nd law in many ways. You can choose to treat all three masses as a single system, which may be the smart thing to do here. Or you can treat each mass separately, making sure to account for the tensions in the strings. (I suspect you'll be asked about those tensions shortly.)

 

[khem student]

Okay that did work at 106 N.
Now to find Tension if the acceleration 1.5 m/s^2!
the total weight is 70.9 kg
so i did (70.9*1.5)+(70.9*9.8)=801.2 N for Tension but got it wrong?

 

[khem student]

oops! yes the total mass is 70.9

 

[khem student]

I also tried (70.90)(1.5+9.8) and got the same answer.
I don't know what I am doing wrong.

 

[Doc Al]

Okay that did work at 106 N.

Good. To find the applied force F, you took all three masses as a single system.

Now to find Tension if the acceleration 1.5 m/s^2!
the total weight is 70.9 kg
so i did (70.9*1.5)+(70.9*9.8)=801.2 N for Tension but got it wrong?

You need to pick a system where the tension is an external force. There are several choices.

For instance: Why not analyze the forces on that third block?

 

[khem student]

it was a rounding error, thank you for your help! I appreciate it!

 

[Doc Al]

it was a rounding error, thank you for your help! I appreciate it!

What was a rounding error? Your answer for the tension was way off!