Solving Tension on Slope w/ 2 Packages

In summary, the question discusses two packages sliding down a ramp with a 30 degree angle and connected by a rope. The packages have different masses and coefficients of kinetic friction. The goal is to find the tension in the rope and the final velocities of the packages. The approach involves summing the forces in the direction of the ramp for each box, but there is confusion over whether the tension can be positive or not.
  • #1
cjavier
17
0
The question: Two packages have just started sliding down a ramp that forms an angle of 30 degrees with the horizontal. They are connected with a rope. Package A is 6m up along the ramp, with a mass of 4 kg and a coefficient of kinetic friction of 0.3. Package B is directly above 4 m further up the ramp, with mass of 8kg and a coefficient of kinetic friction of 0.2. The packages are placed such that if B began moving and A did not, B would have to go through A to move down, i.e they cannot pass each other on the side.

1) Find the tension in the rope just after they start sliding.
2) Find the velocity of both packages when package A arrives at the bottom of the ramp.

I am not sure how to approach this problem. My teacher wrote on my test that T=0 because the rope is slack. He then wrote that the final velocity of package A was 5.5m/s (although I cannot read his handwriting) and that the final velocity of package B was 7.7m/s.

Please help! My approach was to sum the forces in the direction of the ramp for each box like so:

BOX A: ƩFx = mbox-aa = mgsin(30) - T - μkmgcos(30)
BOX B: ƩFx = mbox-ba = mgsin(30) + T - μkmgcos(30)

But I could not derive T becoming 0. Thank you all.
 
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  • #2
If T > 0 then the two accelerations are the same.
a = gsin(30) - T/ma - μkagcos(30)
a = gsin(30) + T/mb - μkbgcos(30)
What do you get if you combine those equations, eliminating a, and use the fact that μka > μkb?
 
  • #3
haruspex said:
If T > 0 then the two accelerations are the same.
a = gsin(30) - T/ma - μkagcos(30)
a = gsin(30) + T/mb - μkbgcos(30)
What do you get if you combine those equations, eliminating a, and use the fact that μka > μkb?

I'm not quite sure what I am supposed to get. I get T/ma + T/mb= -gcos(30)
[I substituted μka with 2 and μkb with 1 for simplicity]
 
  • #4
cjavier said:
I'm not quite sure what I am supposed to get. I get T/ma + T/mb= -gcos(30)
[I substituted μka with 2 and μkb with 1 for simplicity]
So can T be positive?
 
  • #5


I would approach this problem by first identifying all the relevant variables and equations that can be used to solve for the tension in the rope and the velocities of the packages. From the given information, we know the masses, angles, and coefficients of kinetic friction of both packages. We also know that the packages are connected by a rope and that they cannot pass each other on the side.

To solve for the tension in the rope, we can use the equation for Newton's Second Law, which states that the sum of all forces acting on an object is equal to its mass times its acceleration. In this case, we can apply this equation to both packages separately, assuming that they are both moving with the same acceleration down the ramp. This will give us two equations with two unknowns (the tension in the rope and the acceleration of the packages). By solving these equations simultaneously, we can find the value of the tension in the rope.

To solve for the velocities of the packages, we can use the equation for conservation of energy, which states that the initial potential energy of an object is equal to its final kinetic energy. In this case, we can apply this equation to both packages separately and solve for their final velocities.

It is important to note that the rope being slack does not necessarily mean that the tension is 0. The tension in the rope will depend on the forces acting on both packages and can change as they move down the ramp. Additionally, the final velocities of the packages will also depend on the forces acting on them, including the tension in the rope.

In summary, to solve this problem as a scientist, I would use the relevant equations and variables to analyze the forces and energy involved in the motion of the packages. I would also make sure to clearly define any assumptions or limitations in the given information and consider how they may affect the final results.
 

1. How do I calculate the tension on a slope with 2 packages?

To calculate the tension on a slope with 2 packages, you will need to use the formula T = mg(sinθ + μcosθ), where T is the tension, m is the mass of the package, g is the gravitational acceleration, θ is the angle of the slope, and μ is the coefficient of friction. Plug in the values for each package and add them together to get the total tension.

2. What is the coefficient of friction and how do I find it?

The coefficient of friction is a measure of the resistance between two surfaces when they come into contact. To find the coefficient of friction, you will need to perform an experiment where you measure the force required to move one package over the other. The coefficient of friction is equal to the force divided by the weight of the package.

3. Can I use the same formula for different angles of the slope?

Yes, the formula T = mg(sinθ + μcosθ) can be used for any angle of the slope. Just make sure to plug in the correct angle for θ and the corresponding values for m, g, and μ.

4. Is there a maximum tension that can be applied to the packages on the slope?

Yes, there is a maximum tension that can be applied to the packages on the slope. This is determined by the coefficient of friction and the angle of the slope. If the tension exceeds this maximum value, the packages will start to slide down the slope.

5. How can I reduce the tension on the slope with 2 packages?

There are a few ways to reduce the tension on the slope with 2 packages. One way is to reduce the mass of the packages. Another way is to decrease the angle of the slope. You can also decrease the coefficient of friction between the packages. Lastly, you can add additional support or friction pads to the slope to distribute the weight and reduce tension.

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