# Force on two boxes attached by a string

• asdf12321asdf
In summary, the two blocks are being pulled to the right by a 30N force, making an angle of 25o. The coefficient of kinetic friction is 0.100, so the system is experiencing an acceleration of .833m/s². The tension in the string is unknown, but can be estimated to be 4.9 N.
asdf12321asdf

## Homework Statement

Two blocks connected by a light string are being pulled to the right by a constant force of
magnitude 30.0 N making an angle of 25o from the horizontal. The coefficient of kinetic friction
between each block and the surface is 0.100. If m1 = 5.00 kg and m2 = 10.0 kg , (a) Draw a free body diagram for each object. (b) Find the accleration of the system. (c) Find the tension in the string.

F=ma

## The Attempt at a Solution

My attempt at part a:

My attempt at part b:

I got that the force in the x direction of block 2 being pulled is 27.2. So I got that the sum of the forces would be: 27.2-9.8-4.9 = 12.5

Since F=ma and m equals 15, I got that the acceleration equals .833m/s², but that is not the correct answer.

Edit: OK I think I have figured out part b. I realized that the 30N force on m2 was causing a force of 12.7N in the y direction, making the normal force on m2 equal to 85.3N, making the kinetic friction on m2 equal to 8.53N. So why you sum up the forces in the x direction you get: 27.2-8.53-4.9 = 13.77. I then divided by the total weight which is 15 and got .918. Is this correct?

My attempt at part c:
I am really not sure how begin to solve part c

Last edited:
asdf12321asdf said:
Edit: OK I think I have figured out part b. I realized that the 30N force on m2 was causing a force of 12.7N in the y direction, making the normal force on m2 equal to 85.3N, making the kinetic friction on m2 equal to 8.53N. So why you sum up the forces in the x direction you get: 27.2-8.53-4.9 = 13.77. I then divided by the total weight which is 15 and got 9.18. Is this correct?
Yes, all good.

Hint: The tension is an unknown force. In your FBDs, you show the tension force as 4.9 N! Just show it as unknown T. (And, you realize, that your value for the normal force for m2 is incorrect. In the diagram, I'd just show the normal forces as N1 and N2. You'll figure them out as needed.)

My attempt at part c:
I am really not sure how begin to solve part c
Revise your FBDs as suggested above. Use the acceleration to figure out the tension. Hint: Apply Newton's 2nd law to an individual block.

Ok I think I figured out part c. I used F=ma so F=(5)(.918)=4.59. I then added the friction on block 1 and got 9.49N. Is this correct?

And for the free-body diagrams, I don't need any of the actual values? I was never really sure about this so I just filled them in.

asdf12321asdf said:
Ok I think I figured out part c. I used F=ma so F=(5)(.918)=4.59. I then added the friction on block 1 and got 9.49N. Is this correct?
Yes, that's correct. I would describe it like this:
ΣF = ma
T - 4.9 = (5)(.918) = 4.59
T = 9.49 N

And for the free-body diagrams, I don't need any of the actual values? I was never really sure about this so I just filled them in.
The purpose of the free body diagram is to help you analyze forces and solve for the unknowns. (That's why it was the first part of this problem.) If you already knew the values, you'd wouldn't need a diagram--the problem would already be solved.

Oh ok yeah that makes sense.

Thanks for all of your help! I think I am finally starting to understand some of this stuff.

## What is the force on two boxes attached by a string?

The force on two boxes attached by a string is the tension in the string. This tension is equal to the force applied to one box and transmitted through the string to the other box.

## How does the force on two boxes attached by a string change with the angle of the string?

The force on two boxes attached by a string is dependent on the angle of the string. As the angle increases, the force on each box decreases. At 90 degrees, the force on each box is equal to half of the force applied to one box.

## What happens to the force on two boxes attached by a string when one box is heavier than the other?

The force on two boxes attached by a string is still equal to the tension in the string, regardless of the weight of each box. However, the heavier box will experience a greater downward force from its weight, causing the string to stretch and increase the tension.

## Can the force on two boxes attached by a string be greater than the force applied to one box?

No, the force on two boxes attached by a string is always equal to the force applied to one box. This is due to the principle of action and reaction, where the force applied to one box is transmitted through the string to the other box.

## How does the distance between the two boxes affect the force on two boxes attached by a string?

The distance between the two boxes does not directly affect the force on two boxes attached by a string. However, a longer distance may cause the string to stretch more, increasing the tension and therefore the force on each box.

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