Physics I Help needed

[jai6638]

A sled weighing 205 N rests on a 15° incline, held in place by static friction (Figure 5-61). The coefficient of static friction is 0.5. The sled is pulled up the incline at constant speed by a child. The child weighs 500 N and pulls on the rope with a constant force of 100 N. The rope makes an angle of 30° with the incline and has negligible weight. What is the magnitude of the force exerted on the child by the incline?

I said that Fn - mgcos(theta) - Ft ( sin(theta)) = ma = 0
Hence, Fn = mgcos(theta) + FT ( sin(theta)) and got an answer of 532.962N but this is wrong.

What am I doing wrong here?

Thanks much.

 

[TanisHalfElven]

english i help needed too.
will answer later when less busy.

 

[jai6638]

its due at midnight so anytime before that would be appreciated. thanks!

 

[jagec]

(a)if it's moving, static friction is meaningless. WTF?
(b)what kind of piss-poor sled has a coefficient of 0.5? Is the incline covered in sandpaper?
(c)how old is this child? 500N is over 100 lbs. Since they said "child" and not "teenager", I'm going to guess that this kid is fat, and as such probably isn't going to bother pulling the sled for much longer before going inside to watch TV.

 

[oiprocs]

Tsk tsk.

Where's your force of friction in the equation?

 

[jai6638]

So you are saying that it should be

Fn = mgcos(theta) + FT ( sin(theta)) - Fr
Fn = mgcos(theta) + Ft ( sin(theta ) - (Us*mgcos(theta))

??

 

[PottedMeat]

Ugh. Is this the UT physics homework service?

If I was really lazy I'd try 500N first. Sometimes they like to add lots of superfluous info that you have to filter through when the answer is staring at you.

I guess it would really be force of child with gravity + the 45deg portion of the 100N thing.

 

[CraKaJaX]

Originally posted by: jagec
(a)if it's moving, static friction is meaningless. WTF?
(b)what kind of piss-poor sled has a coefficient of 0.5? Is the incline covered in sandpaper?
(c)how old is this child? 500N is over 100 lbs. Since they said "child" and not "teenager", I'm going to guess that this kid is fat, and as such probably isn't going to bother pulling the sled for much longer before going inside to watch TV.

Not necessarily. Is it translational or rotational motion?

 

[jai6638]

its webassign.. pain in the freaking ass. I have only 1 try to go now out of the 5 they give me .. doh

 

[jagec]

Originally posted by: CraKaJaX

Originally posted by: jagec
(a)if it's moving, static friction is meaningless. WTF?
(b)what kind of piss-poor sled has a coefficient of 0.5? Is the incline covered in sandpaper?
(c)how old is this child? 500N is over 100 lbs. Since they said "child" and not "teenager", I'm going to guess that this kid is fat, and as such probably isn't going to bother pulling the sled for much longer before going inside to watch TV.

Not necessarily. Is it translational or rotational motion?

It's a sled. It moves by sliding over the surface of the ground...so having the coefficient of static friction is meaningless.

Of course, based on my calculations the "constant speed" is zero (the 100 N exerted by the child is insufficient to overpower frictional forces and gravity), so the little bastard DEFINITELY is going to give up pretty soon.

That makes the problem much easier, though. Since there is no motion, we have the 500N required to keep the child from falling through the hill (obviously the normal force is somewhat lower because of the slope, but it asks for the *total* force, not just the normal force), and the normal component of the 100N force through the rope. The other component is balanced by the sled.

So, 500 + 100N*sin(theta) is my answer.

/edit: Wait, I'm dumb. The child is stationary. The hill therefore has to balance all forces acting on the child...500+100 N. Think about how you have to dig your heels into the ground when you pull on a rope.

 

[CraKaJaX]

Originally posted by: jagec

Originally posted by: CraKaJaX

Originally posted by: jagec
(a)if it's moving, static friction is meaningless. WTF?
(b)what kind of piss-poor sled has a coefficient of 0.5? Is the incline covered in sandpaper?
(c)how old is this child? 500N is over 100 lbs. Since they said "child" and not "teenager", I'm going to guess that this kid is fat, and as such probably isn't going to bother pulling the sled for much longer before going inside to watch TV.

Not necessarily. Is it translational or rotational motion?

It's a sled. It moves by sliding over the surface of the ground...so having the coefficient of static friction is meaningless.

Of course, based on my calculations the "constant speed" is zero (the 100 N exerted by the child is insufficient to overpower frictional forces and gravity), so the little bastard DEFINITELY is going to give up pretty soon.

That makes the problem much easier, though. Since there is no motion, we have the 500N required to keep the child from falling through the hill (obviously the normal force is somewhat lower because of the slope, but it asks for the *total* force, not just the normal force), and the normal component of the 100N force through the rope. The other component is balanced by the sled.

So, 500 + 100N*sin(theta) is my answer.

I know, I was just being a pain in the ass. But it didn't work on your cheetah-like self.

 

[jai6638]

Originally posted by: jagec

Originally posted by: CraKaJaX

Originally posted by: jagec
(a)if it's moving, static friction is meaningless. WTF?
(b)what kind of piss-poor sled has a coefficient of 0.5? Is the incline covered in sandpaper?
(c)how old is this child? 500N is over 100 lbs. Since they said "child" and not "teenager", I'm going to guess that this kid is fat, and as such probably isn't going to bother pulling the sled for much longer before going inside to watch TV.

Not necessarily. Is it translational or rotational motion?

It's a sled. It moves by sliding over the surface of the ground...so having the coefficient of static friction is meaningless.

Of course, based on my calculations the "constant speed" is zero (the 100 N exerted by the child is insufficient to overpower frictional forces and gravity), so the little bastard DEFINITELY is going to give up pretty soon.

That makes the problem much easier, though. Since there is no motion, we have the 500N required to keep the child from falling through the hill (obviously the normal force is somewhat lower because of the slope, but it asks for the *total* force, not just the normal force), and the normal component of the 100N force through the rope. The other component is balanced by the sled.

So, 500 + 100N*sin(theta) is my answer.

/edit: Wait, I'm dumb. The child is stationary. The hill therefore has to balance all forces acting on the child...500+100 N. Think about how you have to dig your heels into the ground when you pull on a rope.

so the answer is just 600 N? damn.. webassign usually does not give such nice figures as the answers.

 

[jagec]

Originally posted by: jai6638
so the answer is just 600 N? damn.. webassign usually does not give such nice figures as the answers.

That's assuming that they're not really looking for the normal force...wouldn't be the first time that someone screwed up on their phrasing.

 

[CraKaJaX]

Originally posted by: jai6638

Originally posted by: jagec

Originally posted by: CraKaJaX

Originally posted by: jagec
(a)if it's moving, static friction is meaningless. WTF?
(b)what kind of piss-poor sled has a coefficient of 0.5? Is the incline covered in sandpaper?
(c)how old is this child? 500N is over 100 lbs. Since they said "child" and not "teenager", I'm going to guess that this kid is fat, and as such probably isn't going to bother pulling the sled for much longer before going inside to watch TV.

Not necessarily. Is it translational or rotational motion?

It's a sled. It moves by sliding over the surface of the ground...so having the coefficient of static friction is meaningless.

Of course, based on my calculations the "constant speed" is zero (the 100 N exerted by the child is insufficient to overpower frictional forces and gravity), so the little bastard DEFINITELY is going to give up pretty soon.

That makes the problem much easier, though. Since there is no motion, we have the 500N required to keep the child from falling through the hill (obviously the normal force is somewhat lower because of the slope, but it asks for the *total* force, not just the normal force), and the normal component of the 100N force through the rope. The other component is balanced by the sled.

So, 500 + 100N*sin(theta) is my answer.

/edit: Wait, I'm dumb. The child is stationary. The hill therefore has to balance all forces acting on the child...500+100 N. Think about how you have to dig your heels into the ground when you pull on a rope.

so the answer is just 600 N? damn.. webassign usually does not give such nice figures as the answers.

I HATE Webassign. Some of the stuff they make you put in is so ridiculous. If they ask for an equation as an answer, and you put it in the way they DON'T have it programmed into as the answer, it's wrong. :|

Where do you go to school?

 

[Capt Caveman]

PM Fisher or Smackdown, they have PhD's in Physics.

 

[TanisHalfElven]

Originally posted by: PottedMeat
Ugh. Is this the UT physics homework service?

If I was really lazy I'd try 500N first. Sometimes they like to add lots of superfluous info that you have to filter through when the answer is staring at you.

I guess it would really be force of child with gravity + the 45deg portion of the 100N thing.

the UT physics HW service is kicking my ass (though for eng physics 2) so don't say its easy.

 

[TanisHalfElven]

Originally posted by: jai6638
A sled weighing 205 N rests on a 15° incline, held in place by static friction (Figure 5-61). The coefficient of static friction is 0.5. The sled is pulled up the incline at constant speed by a child. The child weighs 500 N and pulls on the rope with a constant force of 100 N. The rope makes an angle of 30° with the incline and has negligible weight. What is the magnitude of the force exerted on the child by the incline?

I said that Fn - mgcos(theta) - Ft ( sin(theta)) = ma = 0
Hence, Fn = mgcos(theta) + FT ( sin(theta)) and got an answer of 532.962N but this is wrong.

What am I doing wrong here?

Thanks much.

that question makes very very little sense. there's no kinetic friction..... wait a second ofcourse not.

you don't need it. find the component at 90 Deg to the slope. add that to the child weight in the same direction component.

on third thought maybe it is a bad formed question consdering there is no way to figure out the force in the rope.