# Force & Free Fall Projectile Homework Help

• davie08
The force of friction is going against the applied force.In summary, a 3.0kg chair at rest is pulled across the floor by a 20N force. If the co-efficient of friction between the floor and chair is 0.25, calculate:a)the net force on the chair. answer=13Nb)the displacement of the chair after 2.0sm answer=8.4mf

## Homework Statement

1)A 3.0kg chair at rest is pulled across the floor by a 20N force.If the co-efficient of friction between the floor and chair is 0.25, calculate:
a)the net force on the chair. answer=13N
b)the displacement of the chair after 2.0sm answer=8.4m

## Homework Equations

Fn=m*a Fn=Fa+Ff Fg=m*g Ff=m*FT(flipped for perpendicular)

V2=V1+a*t d=V1t+1/2a*t^2

## The Attempt at a Solution

I cnat get it started like I am lost with what equation to pick.

First step is to draw a free body diagram. Did you get that far?

After that, I'm kind of confused about why you can't get it started. You already provided the equations. Just start substituting in what you do know, and solve for what you don't.

ya I drew the diagram what's the first equation that I use?

ya I drew the diagram what's the first equation that I use?

$$\Sigma$$F=ma

so is it 20=3.0*a

so is it 20=3.0*a

No. You didn't sum up the forces. You only listed the "pulling" force, ignoring everything else. That's the point of the free body diagram, to list all of the forces and account for them all.

is Fg 29.4 and I forgot but doesn't Fg=F with the perpendicular and you seem like a pretty good person helping everyone out thanks for the help.

is Fg 29.4 and I forgot but doesn't Fg=F with the perpendicular and you seem like a pretty good person helping everyone out thanks for the help.

You're still missing a force. Yes, normal force and weight cancel out. You've got the pulling force, but what else are you missing?

im missing the force of friction right but I got it form Ff=.25*29.4 and it equaled 7.35 is this right?

im missing the force of friction right but I got it form Ff=.25*29.4 and it equaled 7.35 is this right?

In which direction?

if I add 6.67 and 7.35 with the eqn. Fn=Fa+Ff I get 14 but the answer is 13.

well does the direction matter

if I add 6.67 and 7.35 with the eqn. Fn=Fa+Ff I get 14 but the answer is 13.

Where did you get 6.67?

well does the direction matter

Is force a vector or a scalar quantity?

I got 6.67 from 20=3*a but I guess that's wrong right but was the 7.35 right I am not sure what you mean by what direction like the friction is going against the applied force if that what you mean

I got 6.67 from 20=3*a but I guess that's wrong right but was the 7.35 right I am not sure what you mean by what direction like the friction is going against the applied force if that what you mean

Ignore the 6.67, because that equation wasn't correct. You forgot to include the force of friction, which you just identified, and in any case, diving by mass will just give you acceleration. You're not looking for acceleration just yet. You just want the net force.

Now, what I mean by direction is what direction the force is acting. As I'm sure you learned in class, force is a vector quantity. What that means is it has both magnitude and direction. The force of friction always acts to OPPOSE motion, which makes logical sense.

Take a look at the normal force and weight. If you have a book sitting on a table, not moving, the only two forces are the normal force and it's weight. The weight is m*g. The normal force is -(m*g). Note the negative sign. That shows direction. I'm not sure if you're taking calc based physics or not, but there's a trigonometric way to express that. Anyway...

Use the equation $$\Sigma$$F=m*a there. You get m*g + (-m*g) = m*a.

It's not accelerating, so you get mg - mg = 0 which makes sense.

You should be able to apply this formula to your current problem. First, it's asking you to sum up the forces on the chair. You already know the weight and normal force cancel out. You know about the pulling force, you also calculated the force of friction.

Sum them up, and note the direction of the force of friction.

Well wouldn't this be Fn=2.45*7.35 but that doesn't work I am sorry for being such a nuisance but I am not very good with equations. It would probably just work better if you show me the work and then I can try and understand it that way.

oh wait sry

it would be Fn=2.45+7.35 but that wouldn't work either

Where did you get 2.45 from?

I thought you said to put 7.35 over 3

I most certainly did not.

Sum up the forces. I've already told you what all of the forces are, and you've already calculated them all. All you need to do is put them together.

I'm going to bed now. Hope you get it. G'night.

thanks you I am going to fail the test