Force - boxes connected by strings

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The discussion revolves around calculating forces and tensions in a system of three boxes connected by strings, with specific weights and friction coefficients. The maximum force before movement is calculated to be 68.6N, while the applied force for constant speed is initially misunderstood, leading to confusion about tension and acceleration. It is clarified that at constant speed, acceleration is zero, which simplifies the calculations. The net force for acceleration at 2m/s² is determined to be 70N, but the applied force must also account for friction, leading to a revised estimate of 73.5N. The importance of correctly identifying forces and using free body diagrams is emphasized for accurate problem-solving.
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Homework Statement



Three boxes are connected by strings of constant length. (A--B--C--> F) An applied force F acts horizontal in the positive x direction. mc=5kg, mb=2mc, ma=2mb. The coefficients of static and kinetic friction are 0.2 and 0.1 respectively and are the same for each box. The boxes are initially at rest.

1) What is the magnitude of the maximum force that can be applied without causing the boxes to move?

2) The magnitude of the applied force is varied until the boxes are moving to the right at a constant speed of 2m/s. What is the magnitude of the force when the boxes are moving at this constant speed?

3) The magnitude of the applied force is varied until the boxes are moving to the right at a constant speed of 2m/s. What is the magnitude of the tension in the string connecting A and B?

4) The applied force is adjusted so that the boxes accelerate to the right at a rate of 2m/s^2. What is the magnitude of the applied force?

Homework Equations



F=ma

The Attempt at a Solution



mc=5kg
mb=10kg
ma=20kg
mtotal=35kg
us=.2
uk=.1

1) 20*9.8 + 10*9.8 + 5*9.8 = 343
343*.2 = 68.6N

2) and 3) I'm not really understanding. I know F=ma, but a isn't given; v=2m/s. And I don't even know where to begin with tension.

4) F=35*2 = 70N

Thanks for your help in advance.
 
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firework218 said:

Homework Statement



Three boxes are connected by strings of constant length. (A--B--C--> F) An applied force F acts horizontal in the positive x direction. mc=5kg, mb=2mc, ma=2mb. The coefficients of static and kinetic friction are 0.2 and 0.1 respectively and are the same for each box. The boxes are initially at rest.

1) What is the magnitude of the maximum force that can be applied without causing the boxes to move?

2) The magnitude of the applied force is varied until the boxes are moving to the right at a constant speed of 2m/s. What is the magnitude of the force when the boxes are moving at this constant speed?

3) The magnitude of the applied force is varied until the boxes are moving to the right at a constant speed of 2m/s. What is the magnitude of the tension in the string connecting A and B?

4) The applied force is adjusted so that the boxes accelerate to the right at a rate of 2m/s^2. What is the magnitude of the applied force?

Homework Equations



F=ma

The Attempt at a Solution



mc=5kg
mb=10kg
ma=20kg
mtotal=35kg
us=.2
uk=.1

1) 20*9.8 + 10*9.8 + 5*9.8 = 343
343*.2 = 68.6N

2) and 3) I'm not really understanding. I know F=ma, but a isn't given; v=2m/s.
What is the value of a when the blocks are moving at a constant speed?
And I don't even know where to begin with tension.
Draw a Free Body Diagram of Block A, which is moving at constant speed.
4) F=35*2 = 70N
That's the NET force, not the applied force.
Thanks for your help in advance.
Welcome to PF!:smile:
 
"What is the value of a when the blocks are moving at a constant speed?"

a at constant speed = 0m/s^2?

"Draw a Free Body Diagram of Block A, which is moving at constant speed."

fw=20 down, fn=20 up
v=2m/s right, uk=.1 left

that's about all i got.

"That's the NET force, not the applied force."

Fapp= 35*2 + 0.1*35 = 73.5?
 
firework218 said:
"What is the value of a when the blocks are moving at a constant speed?"

a at constant speed = 0m/s^2?
yes, so proceed as you did in part a, except change the friction coefficients.
"Draw a Free Body Diagram of Block A, which is moving at constant speed."

fw=20 down, fn=20 up
v=2m/s right, uk=.1 left

that's about all i got.
fw down is 20*9.8 N, fn up is 20*9.8 N. But what are the forces in the x direction? The velocity v = 2 m/s is not a force, and its value is not relevant in this problem. Only the fact that v is constant is relevant (a = 0).
"That's the NET force, not the applied force."

Fapp= 35*2 + 0.1*35 = 73.5?
Almost correct, you again forgot to convert the total mass to total weight when calculating the friction force.
 
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