Does Pulling or Pushing Reduce Friction More When Moving a Crate?

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SUMMARY

This discussion analyzes the mechanics of moving a crate by comparing the forces required to push it at a downward angle of 30 degrees versus pulling it at an upward angle of 30 degrees. The crate's mass is 50.0 kg, and the coefficient of friction with the concrete is 0.750. It is concluded that pulling the crate results in a lower force requirement due to a reduction in the normal force, thereby decreasing friction. The calculations involve using the equations Fp cos(30°) = μN, with N defined differently for pushing and pulling scenarios.

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Homework Statement


A person has a choice while trying to move a crate across a horizontal pad of concrete: push it at a downward angle of 30 degrees, or pull it at an upward angle of 30 degrees.

If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.

2. The attempt at a solution
I don't know how to start calculating it by I do have an idea of what the answer will be. In the end, pulling at an upward angle will have a lower force because pulling upward will decrease friction resulting in decrease of normal force.

As for the calculation part, I was thinking of using Fpcos(30°)p = \muN with N being (mg + Fpsin(30°)) for pushing and then Fpcos(30°) = \muN with N being (mg - Fpsin(30°)) for pulling but I feel like it's a bit disorganized since it doesn't follow what we usually have in class like a parent formula on top and you can manipulate it to get this and that, etc.
 
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santoki said:

Homework Statement


A person has a choice while trying to move a crate across a horizontal pad of concrete: push it at a downward angle of 30 degrees, or pull it at an upward angle of 30 degrees.

If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.

2. The attempt at a solution
I don't know how to start calculating it by I do have an idea of what the answer will be. In the end, pulling at an upward angle will have a lower force because pulling upward will decrease friction resulting in decrease of normal force.

As for the calculation part, I was thinking of using Fpcos(30°)p = \muN with N being (mg + Fpsin(30°)) for pushing and then Fpcos(30°) = \muN with N being (mg - Fpsin(30°)) for pulling but I feel like it's a bit disorganized since it doesn't follow what we usually have in class like a parent formula on top and you can manipulate it to get this and that, etc.

Start by doing a free body diagram for each of the two scenarios, drawing vectors for all forces acting on the body in each scenario. What is the condition for motion at constant velocity (hint: what must the vectors in each diagram sum to?)?

AM
 

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