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A few Motion Questions

  • Thread starter Paymemoney
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  • #1
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Homework Statement


A 2200kg airplane is pulling two gliders, the first of mass 310kg and the second of mass 260kg, down the runway with an acceleration of 1.9m/s^2. Neglecting the mass of the two ropes and any frictional forces, determine the tension force in the second rope.


Homework Equations


F=ma

The Attempt at a Solution


This is what i have done, however i do not understand the book's answers.

Two forces acting on rope 2, so...

T-494 = 4180

T=4674N

book's answers is 494N, wouldn't this be the net force on the second glider??

Homework Statement


A child pulls an 11kg wagon with a horizontal handle whose mass is 1.8kg, giving the wagon and handle an acceleration of 2.3m/s^2.
a) Find the tension at each end of the handle?

Homework Equations


F=ma

The Attempt at a Solution


Can someone check if my answer is correct.

Net Force of Handle and Wagon are:

F=4.14N and F=25.3N

Tension from handle to wagon

T1-4.14 = 25.3
T1=29.22N

Tension from wagon to handle

25.3-T=4.14
T=20.89N

Homework Statement


In a tractor pulling contest, a 2300kg trctor pulls a 4900kg sledge with an acceleration of 0.61m/s^2. If the tractor exerts a horizontal force of 770N on the ground, determine the magnitude of
a) the force of the tractor on the sledge.
b) the force of the sledge on the tractor and
c) the frictional force exerted on the sledge by the ground.


Homework Equations


F=ma

The Attempt at a Solution


Like the last question can someone tell me if my answer is correct, if not what have i done wrong?

a)Fnet = ma
T - Fnet = ma
T - 7700 = 2300 * 0.61
T - 7700=1403
T=9103N

b) Fnet = ma
T - Fnet = ma
T - 48020 = 4900 * 0.61
T=51009N

c)
T - Friction = ma
9103 - Friction = 4900 * 0.61
Friction = 6114N

P.S
 

Answers and Replies

  • #2
Doc Al
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Homework Statement


A 2200kg airplane is pulling two gliders, the first of mass 310kg and the second of mass 260kg, down the runway with an acceleration of 1.9m/s^2. Neglecting the mass of the two ropes and any frictional forces, determine the tension force in the second rope.


Homework Equations


F=ma

The Attempt at a Solution


This is what i have done, however i do not understand the book's answers.

Two forces acting on rope 2, so...

T-494 = 4180

T=4674N
I don't understand your reasoning. Where do these number come from?

Hint: What forces act on the second glider? Apply Newton's 2nd law.

book's answers is 494N, wouldn't this be the net force on the second glider??
Yes!
 
  • #3
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There is two forces on the second glider and that is Tension and Fnet on glider so Tension is what i want to find.

Fnet of second glider =ma
Fnet of second glider = 260 * 1.9
Fnet of second glider = 494N

Now i know the Fnet i can sub into T-Fnet = ma (force of airplane)

T - 494 = 4180
T=4674N
 
  • #4
Doc Al
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44,880
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There is two forces on the second glider and that is Tension and Fnet on glider so Tension is what i want to find.

Fnet of second glider =ma
Fnet of second glider = 260 * 1.9
Fnet of second glider = 494N

Now i know the Fnet i can sub into T-Fnet = ma (force of airplane)

T - 494 = 4180
T=4674N
Fnet is not a force--it's the sum of all the forces. There's only one horizontal force acting on the second glider--the tension in the rope. (And if you are analyzing the glider, m would be the mass of the glider, not the plane.)
 
  • #5
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but wouldn't there be a force acting from the second glider to first rope


Tension Rope 1<---------glider 2 ------------->Tension Rope 2
 
  • #6
Doc Al
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but wouldn't there be a force acting from the second glider to first rope


Tension Rope 1<---------glider 2 ------------->Tension Rope 2
It depends on which glider is glider2. I took glider2 to be the last one in the chain, like this:

glider2---(rope2)---glider1---(rope1)---plane

So glider2 only has one rope pulling on it.
 
  • #7
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yeh the book's doesn't have labels on the plane or glider it just has it like

glider-----rope---glider----rope---plane

How do you know which glider is which???
 
  • #8
Doc Al
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How do you know which glider is which???
They start counting from the plane.
 
  • #9
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oh ic, i understand now thanks for the help.
 
  • #10
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Can anyone answer my other question.
 
  • #11
Doc Al
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A child pulls an 11kg wagon with a horizontal handle whose mass is 1.8kg, giving the wagon and handle an acceleration of 2.3m/s^2.
a) Find the tension at each end of the handle?

Homework Equations


F=ma

The Attempt at a Solution


Can someone check if my answer is correct.

Net Force of Handle and Wagon are:

F=4.14N and F=25.3N

Tension from handle to wagon

T1-4.14 = 25.3
T1=29.22N

Tension from wagon to handle

25.3-T=4.14
T=20.89N
Not quite right. And your method is a bit confusing. The handle has two ends: the child end and the wagon end. What force does the child exert on the handle? That's the tension at the child end. What force does the wagon exert on the handle? That's the tension at that end. (Hint: Newton's 3rd law.)
 
  • #12
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Well, i calculated that if there is only one force on each of the object, then logically speaking the tension on both ends are equal and opposite according to Newton's Third Law.

However if the masses are different on each object would the force exerted be different or the same?

If they are different would the Tension at the child's end be 4.14N and the wagon's end be 25.3N?
 
  • #13
Doc Al
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Which exerts the greater force: the child on the handle, or the handle on the wagon?

It may help to draw a force diagram showing all the forces acting on the wagon and on the handle.
 
  • #14
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The child exerts one force

child--------->handle

Handle exerts two forces, one on the child and the another one on the wagon

child<--------handle--------->wagon

To answer your question i would say the handle on the wagon exert more force.
 
  • #15
Doc Al
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The child exerts one force

child--------->handle

Handle exerts two forces, one on the child and the another one on the wagon

child<--------handle--------->wagon
Good.

To answer your question i would say the handle on the wagon exert more force.
No. The child must exert enough force to accelerate both wagon + handle, while the handle only needs to exert enough force on the wagon to accelerate the wagon.

(wagon + handle)---> child

(wagon)---> handle
 
  • #16
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so the child will exert the total force of the wagon and handle, which is 4.14 + 25.3 = 29.44N

so the wagon end will exert 25.3N
 
  • #17
Doc Al
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so the child will exert the total force of the wagon and handle, which is 4.14 + 25.3 = 29.44N

so the wagon end will exert 25.3N
Right!
 
  • #18
Doc Al
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In a tractor pulling contest, a 2300kg trctor pulls a 4900kg sledge with an acceleration of 0.61m/s^2. If the tractor exerts a horizontal force of 770N on the ground, determine the magnitude of
a) the force of the tractor on the sledge.
b) the force of the sledge on the tractor and
c) the frictional force exerted on the sledge by the ground.


Homework Equations


F=ma

The Attempt at a Solution


Like the last question can someone tell me if my answer is correct, if not what have i done wrong?

a)Fnet = ma
T - Fnet = ma
T - 7700 = 2300 * 0.61
T - 7700=1403
T=9103N
You messed up the direction (and thus the sign) of the forces. (Also: Don't mix up an individual force--the 7700N--with the net force.)
 
  • #19
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so the tractor on the sedge would exert two forces

sledge<--------tractor--------->7700N

7700-T = 2300 * 0.61
T=6297N

The sledge exerts one force on the tractor

sledge------>

F=ma
F=4900*0.61
F=2989N
 
  • #20
Doc Al
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so the tractor on the sedge would exert two forces

T<--------tractor--------->7700N
I'd phrase that this way: There are two forces acting on the tractor:
(1) The force from the sledge, called T.
(2) The force from the ground, which we know is 7700 N.

7700-T = 2300 * 0.61
T=6297N
Good.

The sledge exerts one force on the tractor

sledge------>

F=ma
F=4900*0.61
F=2989N
That's the net force on the sledge. To answer (b), consider Newton's 3rd law.
 
  • #21
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so if the tractor exerts 6297N on the sledge then according to Newton's 3rd Law it will push back with 6297N but in the opposite direction.
 
  • #22
Doc Al
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so if the tractor exerts 6297N on the sledge then according to Newton's 3rd Law it will push back with 6297N but in the opposite direction.
If the tractor exerts 6297N on the sledge, then the sledge exerts 6297N on the tractor (in the opposite direction, of course).
 
  • #23
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so for question c) the frictional force would be:
6297-Friction = 2989

Friction = 3308N
 
  • #24
Doc Al
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so for question c) the frictional force would be:
6297-Friction = 2989

Friction = 3308N
Right!
 

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