# Did I Calculate the Tension in the Cord Correctly?

• apiwowar
In summary, the conversation discusses a scenario with two boxes connected by a cord on a frictionless incline. The question asks for the tension in the cord with a given horizontal force, and the largest value of the force without the cord becoming slack. A mistake is pointed out in the calculation of the tension, as the weight of the second box on the incline should also be considered. The conversation ends with a hint to consider the weight parallel to the surface.
apiwowar
Figure 5-56 shows a box of mass m2 = 1.3 kg on a frictionless plane inclined at angle θ = 34°. It is connected by a cord of negligible mass to a box of mass m1 = 2.7 kg on a horizontal frictionless surface. The pulley is frictionless and massless. (a) If the magnitude of the horizontal force F is 2.5 N, what is the tension in the connecting cord? (b) What is the largest value the magnitude of F may have without the connecting cord becoming slack?

so i broke this up into 2 free body diagrams

for the first one i got T+2.5 = 2.7a so T = 2.7a-2.5

for the second one T-7.1 = 1.3a so T=1.3a+7.1

putting them together and solving for a i got a = 6.86m/s^2

when i plugged that back to solve to T i got T = 16.0N

but its ssaying that 16.0 is wrong, did i do anything wrong?

and can i get a hint for part b?

You forgot to take into account the fact that it's only the vertical component of the weight that's going to affect the tension.

what do you mean?

Well, imagine if that 34 degree incline was non existent, and the m2 was instead hanging in the air, with no surface supporting it. Don't you think that would change the tension in the cable?

That value of 16.0N that you calculated would be correct if in such a scenario because you haven't taken into account that there is some support being provided to m2.

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if there was no surface then the force downward would just be the weight, but since there is a surface you find the weight that is parallel to that surface which is mgsin(x). is that what youre getting at?

Yeah, whether it's sin or cos I'm not sure, but since you have the final answer already you can figure that one out.

## 1. How does a frictionless pulley and incline work?

A frictionless pulley and incline system involves a pulley and an inclined plane, where the pulley is attached to a load and the inclined plane is used to lift the load. The lack of friction allows for the load to be lifted with minimal effort.

## 2. What is the purpose of using a frictionless pulley and incline?

The purpose of using a frictionless pulley and incline is to reduce the amount of force needed to lift an object. This can make tasks easier and more efficient.

## 3. How does friction affect the efficiency of a pulley and incline system?

In a frictionless system, there is no loss of energy due to friction. However, in a real-world system with friction, some energy is lost due to the resistance between the moving parts. This can decrease the efficiency of the system.

## 4. What is the difference between a frictionless pulley and incline and a regular one?

A frictionless pulley and incline system has no friction, which means that the force needed to lift an object is equal to the weight of the object. In a regular system with friction, the force needed is greater than the weight of the object due to the energy lost to friction.

## 5. Can a frictionless pulley and incline system be used in real-world applications?

While it is not possible to achieve a truly frictionless system, friction can be minimized through the use of materials and lubricants. This allows for a more efficient pulley and incline system to be used in real-world applications, such as in elevators and cranes.

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