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**Rod is 1. pulled, 2. pushed horizontally. Force required to sustain constant vel.???**

## Homework Statement

There are 2 scenarios.

1. A rod of mass 5kg is first pulled at constant velocity by a force at 45° to the horizontal.

2. Then the same rod is pushed at constant velocity by a force at 45° to the horizontal.

Assuming that in both cases the frictional force is horizontal and equal to 0.4 times the normal reaction force on the rod, find the force F in each case.

Why is the force different in 1 and 2? (Especially important to me, because I dont understand why it should be different)

## Homework Equations

weight = mass x gravity

normal reaction force N = cos(x) x weight

friction force f = 0.4 x N

f = x-component of F. therefore.. F = f/ cos (x)

## The Attempt at a Solution

In scenario 1:

I found the weight to be 50N.

Then the reaction force must be cos (45) x 50N = 35N.

Thus the friction must be 0.4 x 35 N = 14 N

Thus F = 14 / cos (45) =19.71 N

This answer is wrong (F = 20.2 N according to answer key) and I don't understand why.

In scenario 2, I don't know why it makes a difference if the rod is pushed or pulled. I again i get F = 19.71 N, even though the answer is supposed to be F = 47.1 N

Thank you for your support!