# Newtonian motion

1. The problem statement, all variables and given/known

You are playing with your daugther in the snow. She sits on a sled and asks you to slide her across a flat , horizontal field. You have a choice of (a) pushing her from behind by applying a force downward on her shoulders at 30 degrees below the horizontal (b) attaching a rope to the front of the sled and pulling with a force 30 degrees above the horizontal. Which would be easier for you and why

2. Homework Equations
F =ma

## The Attempt at a Solution

I think both situations yield the same net force thus the same acceleration

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mfb
Mentor
I think both situations yield the same net force thus the same acceleration
Consider the case where the sled has a constant velocity. Do you still need a force to keep it moving? What about that effect (it is also relevant while still accelerating)?

Both options are not in constant velocity. Both options have one force in the horizontal direction

mfb
Mentor
Both options are not in constant velocity.
It does not matter.

There is an effect you missed so far. The force won't all be used for acceleration. There is something else to consider.
Hint: it is snow, not ice.

Friction

mfb
Mentor
Correct. How do you calculate the friction force? Does pushing/pulling at an angle change it?

I think for both cases it will be -Fcos@

mfb
Mentor
What is F, and where does the sign come from, and in which direction?

F is the applied force

The sign is a mistake

nrqed
Homework Helper
Gold Member
F is the applied force
I think that what mfb wants to make you think about is the definition of the friction force. What is the fundamental equation that gives the kinetic friction force on a sliding object?

Last edited:
• mfb
Fk = ukN

nrqed
Homework Helper
Gold Member
Fk = ukN
Right. So what can we say about the normal force in the first case compared to the second case?

It has a component in one case?

mfb
Mentor
What does "it has a component" mean? All forces can always be split into separate components. But where do those components point to?

I mean the force is at an angle so it will result in a horizontal and vertical comp.. but the thing is both forces are at angles. But the components of both forces have different direction

Are you trying to point that?

mfb
Mentor
But the components of both forces have different direction
Right.

The y comp of first case will be up while down in second case....so maybe more friction in second case? Is that right

mfb
Mentor
The second case is you pulling up. You expect more force on the ground there?

The case in which you push

There will be a higher normal force....but that doesnt make sense because the will be a higher normal for force in the other case too as Efy=O

mfb
Mentor
What is Efy?

If you push, you have a higher normal force, correct. This leads to larger friction.
If you pull upwards (at an angle), you reduce the normal force.

Sum of forces in the y direction