# Three identical blocks connected by ideal strings, Find Horizontal F

1. May 20, 2013

### khem student

1. The problem statement, all variables and given/known data

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 dont understand is that the blocks are listed as Identical but all vary in weight.

2. Relevant equations

Image is attached.

3. 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...

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2. May 20, 2013

### Staff: Mentor

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

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

Not sure why you think that.

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

3. May 20, 2013

### Staff: Mentor

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

4. May 20, 2013

### 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?

5. May 20, 2013

### Staff: Mentor

Exactly.

6. May 20, 2013

### 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?

7. May 20, 2013

### Staff: Mentor

Good.

By using Newton's 2nd law.

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

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.)

8. May 20, 2013

### 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?

9. May 20, 2013

### khem student

oops! yes the total mass is 70.9

10. May 20, 2013

### khem student

I also tried (70.90)(1.5+9.8) and got the same answer.
I dont know what I am doing wrong.

11. May 20, 2013

### Staff: Mentor

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

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?

12. May 20, 2013

### khem student

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

13. May 20, 2013

### Staff: Mentor

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