Force - boxes connected by strings

[firework218]

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.

 

[PhanthomJay]

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!

 

[firework218]

"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?

 

[PhanthomJay]

"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.

 

[rwisz]

Hello, according to the problem, the boxes are initially at rest and are connected by strings of constant length. In this situation, the forces acting on the boxes are the applied force F, the weight of each box (mg), and the tension in the strings connecting the boxes. We can use the equation F=ma to analyze the motion of the boxes.

1) To determine the maximum force that can be applied without causing the boxes to move, we need to consider the forces acting on the boxes. Since the boxes are initially at rest, the net force acting on the system must be zero. This means that the applied force must be equal in magnitude and opposite in direction to the sum of the weight and tension in the strings. So, we can write the following equation:

F - (20*9.8 + 10*9.8 + 5*9.8) - T = 0

where T is the tension in the strings. The maximum applied force can be found by setting the tension to zero, since this would mean that the strings are not exerting any force on the boxes. So, the maximum applied force is equal to:

Fmax = 20*9.8 + 10*9.8 + 5*9.8 = 343 N

2) and 3) The magnitude of the applied force can be varied until the boxes are moving at a constant speed of 2m/s. This means that the net force acting on the system is zero (since the boxes are not accelerating). We can use the same equation as in part 1, but now we know that the tension in the strings is not zero. So, we can write:

F - (20*9.8 + 10*9.8 + 5*9.8) - T = 0

Since the boxes are moving at a constant speed, we know that the net force is equal to the frictional force, which is given by Ff=ukN, where N is the normal force (equal to the weight of the box). So, we can write:

Ff = ukN = uk(20*9.8 + 10*9.8 + 5*9.8) = 68.6N

But we also know that the frictional force is equal to the tension in the strings, so we can set Ff=T and solve for T:

T =