Pushing crates up a frictionless ramp

In summary: I'll email my professor now.In summary, Mario has added two hard nylon rails to the company's standard ramp, which will give the mover great traction as he or she pushes the crates along the rails and up the ramp. Additionally, he has installed a special super-grip carpet in the space between the nylon rails, which will give the mover great traction as he or she pushes the crates along the rails and up the ramp.
  • #1
drjohn15
13
0
Mario's new additions to the company's standard ramp include two hard nylon rails which run the length of the ramp; the crates which have to be moved will slide along the hard nylon rails with almost no friction. In the space on the ramp which is in between the nylon rails, Mario has installed a special super-grip carpet, which will give the mover great traction as he or she pushes the crates along the rails and up the ramp.

They begin with the ramp in a North-South orientation flat on the lab floor, and they put two crates of equal mass side-by-side at the south end of the ramp. They then use two hydraulic jacks to raise the north end of the ramp to a desired height; one crate is then at the bottom of the ramp (where the ramp intersects with the lab floor) and the other crate is adjacent to the first and just a little higher up along the ramp. We will now refer to the first crate as the "lower crate" and to the second crate as the "upper crate".

Mario knows that most of the company movers will use the ramp to push crates at constant speed, but he wishes to test extreme possibilities, so he plans to push the crates up the ramp with increasing speed. He selects two crates, each of mass 47 kg. The start of his push will be at the bottom of the ramp, and he puts a piece of tape at a point 5.7 m along the ramp (not along the floor) which will mark the endpoint of his hard push; upon arriving at this endpoint, he will "ease up" so that the crates will move at constant speed for the rest of their journey the top of the ramp. Luigi measures the height from the lab floor to the marked endpoint as being 250.8 cm. Luigi also designs a system of two electric eyes connected to a digital timer which will time Mario during his trip from the starting point to the marked endpoint. The size of the selected crates is just right so that Mario will push HORIZONTALLY (parallel to Earth's surface) on the lower crate.

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So I know this is lengthy, but I have about 27 questions that have to be answered about this scenario. I'm really just having trouble visualizing this and drawing the free body diagrams.

Any help would be really great! I just need a push in the right direction.

Thanks!
 
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  • #2
So far I have two different scenarios drawn. The first is where the lower crate is pushed by Mario, and the upper crate gets pushed by the lower crate (one in front of the other).

My other drawings show the lower crate at the bottom of the ramp, and the upper crate to the east of the lower one (side by side) and slightly higher up the ramp.
 
  • #3
drjohn15 said:
They begin with the ramp in a North-South orientation flat on the lab floor, and they put two crates of equal mass side-by-side at the south end of the ramp. They then use two hydraulic jacks to raise the north end of the ramp to a desired height; one crate is then at the bottom of the ramp (where the ramp intersects with the lab floor) and the other crate is adjacent to the first and just a little higher up along the ramp. We will now refer to the first crate as the "lower crate" and to the second crate as the "upper crate".

drjohn15 said:
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So I know this is lengthy, but I have about 27 questions that have to be answered about this scenario. I'm really just having trouble visualizing this and drawing the free body diagrams.

I don't blame you, the explanation of the positions of the crates is pretty ambiguous. You may email your professor and ask him/her to clarify what is meant by "adjacent to the first and just a little higher up along the ramp."

However based on context I'm lead to assume that it means basically "resting" directly in front of the first crate.

Further, it might help to see the questions you need answered, perhaps a picture so that you don't have to type it all out.
 
  • #4
Here are some of the questions asked.

***EDIT, I guessed on the first three questions... ****
 

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  • #5
I attached a picture if what you're diagram SHOULD be. Don't take it as absolute though, I'm still not 100% on the intent here.

The diagrams below are individual diagrams of the boxes, and note that I didn't include all the relevant forces nor did I draw the two boxes as a single object, which you should do.

Hopefully this can get you going!

Let us know if you have anymore specific questions concerning the specific solutions.
 

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  • #6
Awesome! Thanks a lot Jesse!
 
  • #7
So I managed to find all of the quantities for the crates at constant speed. But now their throwing in some acceleration and I'm kind of lost as to how to approach this.

I've attached some of the questions that have already been asked and also the questions that I'm having trouble with.

Thanks!
 

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  • #8
drjohn15: how did you find the parts of the first part I am stuck on this part now?
 
  • #9
Apologies for the delayed response. In having trouble viewing those questions but you asked about acceleration, so I can say this: F=MA, if you've found the force mass is given, acceleration easily follows. Just keep in mind the force and acceleration are vectors and the acceleration is ALWAYS in the direction of the Force and vice versa. I'll take a better look at the questoons later this evening and have a more specified response if you need it.
 
  • #10
I know this isn't my post but I have the exact same problem just with different numbers...

How do you find the normal force on the lower crate since it isn't the same as the upper crate because of Mario pushing on the crate.

I took Mario's pushing force and subtracted the force on the lower crate by the upper crate and that gave me (I think) the x component of the normal force.

Am I on the right track?
 

Related to Pushing crates up a frictionless ramp

What is meant by pushing crates up a frictionless ramp?

Pushing crates up a frictionless ramp refers to the act of moving crates along a ramp that has no friction, meaning there is no resistance to motion. This scenario is often used in physics experiments to study the effects of forces on objects in motion.

How does the absence of friction affect pushing crates up a ramp?

The absence of friction allows for the crates to move along the ramp with minimal effort. This is because without friction, there is no force acting in the opposite direction of the motion, making it easier to push the crates up the ramp.

What are the factors that affect pushing crates up a frictionless ramp?

The main factor that affects pushing crates up a frictionless ramp is the force applied to the crates. Other factors that may have an impact include the mass of the crates, the angle of the ramp, and the acceleration due to gravity.

What is the equation for calculating the force required to push crates up a frictionless ramp?

The equation for calculating the force required to push crates up a frictionless ramp is F = mgsinθ, where F is the force, m is the mass of the crates, g is the acceleration due to gravity, and θ is the angle of the ramp.

What are some real-life applications of pushing crates up a frictionless ramp?

Pushing crates up a frictionless ramp has several real-life applications, such as in the design of machinery and vehicles that require minimal friction for efficient movement. It is also used in the study of inclined planes and pulley systems, as well as in the development of sports equipment like skateboards and snowboards.

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