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

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!

## Answers and Replies

Basically what you forgot, the friction depends on the normal force, but the normal force is not only the gravitational force.
It also depends on the vertical component of your applied force.
Thats also where the difference will come from, if you pull the normal force will decrease so will the friction, if you push the normal force will increase and so will the friction.

thank you. that helped.

no problem ;)