# Force - boxes connected by strings

• firework218
In summary, the maximum force that can be applied without causing the boxes to move is 68.6N. When the boxes are moving at a constant speed of 2m/s, the magnitude of the force is 68.6N. The magnitude of tension in the string connecting A and B is 68.6N. To accelerate the boxes to the right at a rate of 2m/s^2, the applied force must be 73.5N."
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?

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

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?

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.
Welcome to PF!

"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.

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 =

## What is force and how is it related to boxes connected by strings?

Force is a physical quantity that describes the interaction between objects. In the case of boxes connected by strings, force is the tension in the strings that connects the boxes together. This force is responsible for keeping the boxes together and determining their motion.

## How do you calculate the force in a string connecting two boxes?

The force in a string connecting two boxes can be calculated using the formula F = ma, where F is the force, m is the mass of the box, and a is the acceleration of the box. If the boxes are connected in a horizontal direction, the force would be equal to the mass of the box times the acceleration of the box. If the boxes are connected in a vertical direction, the force would be equal to the mass of the box times the acceleration due to gravity.

## What happens to the force in the string when one box is heavier than the other?

If one box is heavier than the other, the force in the string connecting them will be greater on the side of the heavier box. This is because the heavier box will have a greater mass, and according to the formula F=ma, the force is directly proportional to the mass. Therefore, the heavier box will experience a greater force in the string compared to the lighter box.

## How does the angle of the string affect the force between the boxes?

The angle of the string does not affect the force between the boxes as long as the string is taut. This is because the force in the string is determined by the tension in the string, which only depends on the mass and acceleration of the boxes. However, if the string becomes slack, the force will decrease as the angle increases due to a decrease in tension.

## Can the force in a string connecting two boxes ever be greater than the weight of the boxes?

No, the force in a string connecting two boxes can never be greater than the weight of the boxes. This is because the tension in the string is always equal to the weight of the boxes in a stable system. If the force in the string were to exceed the weight of the boxes, the system would no longer be in equilibrium and the boxes would move.

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