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Force applied to a wagon [Answer Check]

  • Thread starter lamerali
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  • #1
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Hi, I am taking a correspondence physics course and I've come across a question that I'm not too sure about; I've completed the question but I'm not certain I did it correctly if anyone could please check my answer and correct me if I'm wrong I would really appreciate it!

Question:
In many neighborhoods, you might see parents pulling youngsters in a four-wheeled wagon. The child and the wagon have a combined mass of 50 kg and the adult does 2.2 x 10^3 J of work pulling the two 60 m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.
a) Determine the magnitude of the force applied.
b) Determine the angle at which the parent is applying this force.

My answer:
a)
Fapp = Ff + Fg
= uFn + Fg
=u(mg) + mg
=(0.26)(50 kg)(9.8N/kg) + (50 kg)(9.8 N/kg)
=617.4 N
Therefore the force that the parent applies to the wagon is 617.4 N

b) W = Fapp x cos(theta) x (delta)d
cos(theta) = W / (Fapp(delta)d)
= (2.2 x 10^3 J) / (617.4 x 60 m)
theta = 87 degrees
Therefore the parent is applying this force at 87 degrees to the horizontal.

Thanks in advance!
 

Answers and Replies

  • #2
tiny-tim
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Welcome to PF!

Fapp = Ff + Fg
= uFn + Fg
=u(mg) + mg
Hi lamerali ! Welcome to PF! :smile:

No … forces are vectors, and they add like vectors.

You have added them like scalars (ordinary numbers), as if they were in line.

The µmg force is horizontal, but the mg force is vertical.

Take horizontal components, and start again! :smile:
 
  • #3
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Hi Tim, thanks!

How would I find the horizontal component of mg if they havent given any angles??? :S
 
  • #4
tiny-tim
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How would I find the horizontal component of mg if they havent given any angles??? :S
aha … you're expecting everything to come out in one line, and maths ain't like that. :cry:

The angle is another unknown, just like F … call it θ, and work out both the horizontal and vertical components of forces in terms of θ.

That gives you two equations in your two unknowns (F and θ), which enables you to solve for both. :smile:
 
  • #5
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alright how does this sound...
the vertical component of the force is Fg = mg while the horizontal component is Ff = u(mg) therefore the angle of elevation would be

(theta) = tan^-1((mg)/u(mg))
= 75.4 degrees
and then you can find the applied force by saying
Fapp = (mg) / sin(theta)

I don't know where i'm going with this but thought i'd give it a try :C
 
  • #6
tiny-tim
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Nooo …

Have you drawn a diagram of all the forces?

Because you've left out the reaction force, which wil be less than mg, because the adult is pulling partly upward.

Hint: use the given work.

Try again! :smile:
 
  • #7
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W = Fapp(delta)d
Fapp = w/(delta)d
= (2.2x10^3)/(60)
= 36.6 N
Is that it?? Thats what i had in the beginning but i thought i had to account for the frictional force????
 
  • #8
tiny-tim
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W = Fapp(delta)d
Fapp = w/(delta)d
= (2.2x10^3)/(60)
= 36.6 N
Is that it?? Thats what i had in the beginning but i thought i had to account for the frictional force????
Yes, that's the horizontal component of the force. :smile:

From that, you find the friction force.

From that, you find the normal force.

From that, you find the vertical component of the force. :smile:
 
  • #9
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alright this is what I got I THINK it works

Ff = u(mg)
= 127.4
Fn = mg
= 490

(theta) = tan^-1 (490/127.2)
= 75.4 degrees

Therefore the vertical component of the force is 474.5 N

Then to find the total force you can use the Pythagorean theorem and I got 476 N and the angle can be found using W = Fapp x cos (theta) x (delta)d or any of the trigonometric ratios which is 85.6 degrees

Therefore the parent applies a force of 474.5 N 85.6 degrees to the horizontal???
 
  • #10
tiny-tim
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alright this is what I got I THINK it works
NO IT DOESN'T! :frown:

You've gone back to the method in your post #5.

Why did you abandon your result from you post #7, in which you found that the horizontal component of the applied force was 36.6N?

Hint: using good ol' Newton's second law for horizontal components, if the applied force is 36.6N, what is the friction force? :smile:
 
  • #11
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alright isnt the force of friction just
Ff = u(mg)

that is all my book gives me...this course is really poorly structured they dont explain anything i'm totally LOST :(
 
  • #12
tiny-tim
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alright isnt the force of friction just
Ff = u(mg)

that is all my book gives me...this course is really poorly structured they dont explain anything i'm totally LOST :(
ok … now I see why you're going wrong!

The force of friction is uN, where N is the normal force (the normal component of the reaction force).

If nobody is pulling the wagon, or if the pulling is horizontal, then N = mg, and the book's equation is correct.

But the adult here is pulling the wagon slightly upward, so N < mg, and the book's equation is wrong. :frown:

In this case, the equation Fr = uN still applies …

BUT you don't use it to work out Fr from N, you use it the other way round, to work out N from Fr! :biggrin:

You know Fapp = 36.6, so what is Fr? :smile:
 
  • #13
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Does Fr stand for reaction force? my text has not mentioned anything about Fr at all :| i have no idea what relation Fr has to the other formulas (the only reason why i know such a force exists is because you posted something about it in #6)...these ILC courses expect you to just know this stuff!!! ERRR!!! :C
 
  • #14
tiny-tim
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No … my Fr = uN meant the Friction force equals u times the Normal component of the reaction force.
… my text has not mentioned anything about Fr at all :| i have no idea what relation Fr has to the other formulas (the only reason why i know such a force exists is because you posted something about it in #6)...these ILC courses expect you to just know this stuff!!! ERRR!!! :C
(what's ILC?)
Have you drawn a diagram of all the forces?

Because you've left out the reaction force, which wil be less than mg, because the adult is pulling partly upward.
There is a reaction force between two bodies which are in contact.

Its strength and direction are usually unknown.

To find other parts of the problem, one often has to find the reaction force first.

The "horizontal" component of the reaction force is called the friction force.

The "vertical", or normal, component of the reaction force is what I call N. It is equal and opposite to the sum of "vertical" components of all the other forces.

Always draw it into any diagram.

And, as you already knew, the friction force (the "horizontal" component of the reaction force) is equal and opposite to the sum of "horizontal" component of all the other forces (plus mass times any acceleration).

ok, you know Fapp = 36.6, so what is Ff?​
 
  • #15
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equal but opposite to Fapp would just be Ff = -36.6 N ?????
 
  • #16
tiny-tim
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equal but opposite to Fapp would just be Ff = -36.6 N ?????
Yes … but why all the question marks?

Sometimes maths really is obvious and easy!!!!! :smile:

ok, now what is the normal force?

(Hint: use u).
 
  • #17
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YAY! Thank You! (I hate when that happens the answer is right under you nose but it takes forever to figure it out)
so...

Ff = uN
N = Ff/u
N = (-36.6)/(0.26)
N = -140.8 N
 
  • #18
tiny-tim
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YAY! Thank You! (I hate when that happens the answer is right under you nose but it takes forever to figure it out)
Yes … sometimes it takes half-an-hour to realise that something's obvious! :smile:
Ff = uN
N = Ff/u
N = (-36.6)/(0.26)
N = -140.8 N
ok, the vertical force which the ground exerts on the wagon is 140.8N.

So the vertical component of the pulling force is … ? :smile:
 
  • #19
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would the vertical component of the pulling force be

Fnet = N + Fg
= -140.8 N + (50 kg)(9.8 N/ kg)
= 349.2 N
 
  • #20
tiny-tim
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would the vertical component of the pulling force be

Fnet = N + Fg
= -140.8 N + (50 kg)(9.8 N/ kg)
= 349.2 N
Yes, that looks right. :smile:

So the horizontal component of the pulling force is 36.6N, and the vertical component is 349.2N.

hmm … that doesn't look realistic, does it? … but I can't see anywhere we've gone wrong. :confused:
 
  • #21
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A lot of the questions in my book are not realistic...so if we haven't gone wrong in our calculations that it is probably correct...
 
  • #22
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so now do i do the whole Pythagorean theorem/trigonometric ratio thing that I've been trying to do since the beginning? :D
 
  • #23
tiny-tim
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so now do i do the whole Pythagorean theorem/trigonometric ratio thing that I've been trying to do since the beginning? :D
Yes! :smile:
 
  • #24
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YAY! we did it!! THANKS! I think i can take it from here!!! THANK YOU SOOO MUCH!!! really appreciate all the help!

peace
 

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