Question

A man, using a 70kg garden roller on a level surface, exerts a force of 200N at 45 degrees to the ground. Find the vertical force of the roller on the ground:
A) if he pulls;
B) if he pushes the roller

Answer 1

to save time, you should solve it. I don't want to waste your time by asking a nonsense question @ybarrap

Answer 2

In both cases, the force is downward with 200N. The difference between pulling and pushing is in the tangential component of this force. The resulting force (200N) will be the same as will the magnitude of the normal and tangential components. Since, whether you are pulling or pushing does not change the magnitudes, just direction, the Normal component of this force (i.e. "the vertical force of the roller on the ground"), will be the same.

At 45 degrees, this turns out to be \(200 \sin 45^\circ\) plus the weight of the roller \(m\times g \times\sin 45^\circ=70,000\times 9.81\times \sin 45^\circ \). Sum these two components to get your final answer.

Answer 3

pic, please

Answer 4

n both cases, the force is downward with 200N. The difference between pulling and pushing is in the tangential component of this force. The resulting force (200N) will be the same as will the magnitude of the normal and tangential components. Since, whether you are pulling or pushing does not change the magnitudes, just direction, the Normal component of this force (i.e. "the vertical force of the roller on the ground"), will be the same.

At 45 degrees, this turns out to be \(200 \times \sin 45^∘\) plus the weight of the roller \(m×g=70×9.81\). Sum these two components to get your final answer.



Does this make sense?

Answer 5

I corrected normal component of rake force

Answer 6

yea, what I want to confirm is the force 200N is a hyp leg or \(F_y\)

Answer 7

you know, my English is so bad. I want to know if the problem says so, I should understand it correctly.

Answer 8

\(F_T\) is to the right when pushing the rake (same magnitude), and to the left (same magnitude) when pulling the rake, but \(F_N\) stays the same (magnitude and direction. Note: \(F_N\) the way I've drawn it will be actually negative.

Answer 9

the question just ask for vertical force, why do we have to calculate horizontal one?

Answer 10

We don't need that. That's free. But it does show that the total forces do not change.

Answer 11

ah, so if he pushes, the total force to the ground is \(F_g ~~ and ~~ F_y\) when he pulls , the force is just \(F_g\) right?

Answer 12

No, total force on the ground is Fg and Fy either way.

Answer 13

why? is it not that when he pulls , the force of the machine is off?

Answer 14

It's a roller, not a rake. And he is always pushing down with 200 N at 45 degrees so whether he's pulling or pushing doesn't matter, only the direction changes.

For example, if someone pushes down on a ball at 45 degrees, it doesn't matter which direction the ball was pushed down, \(F_y\) and \(F_g\) will be the same.

Answer 15

got you for this. so if the degree is 90 , F = F_g, at that time, push and pull are different, right?

Answer 16

*F_n = F_g

Answer 17

No. It says in the problem that a force of 200 N was exerted on the ground. That means he is pushing DOWN at ALL times, whether pushing OR pulling.

If the user is above the roller and the problem says the user is "exerting a force" of 200 N on the ground, that means he is pushing down NOT picking up and is independent of whether he is pushing or pulling.

Answer 18

The information about pulling or pushing only affects the tangential force (i.e. direction). It does not affect the Normal component of the force "exerted."

Answer 19

got you.

Answer 20

This language can be very confusing, I know.

Answer 21

thank you.