Tractor Pulling Contest: Determining Forces and Acceleration

• Paymemoney
In summary: The child exerts the greater force. They exert the same force on each other, but the child must accelerate both. The handle only accelerates the wagon.

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.

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?

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.

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

Paymemoney said:

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.

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!

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

Paymemoney said:
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.)

but wouldn't there be a force acting from the second glider to first ropeTension Rope 1<---------glider 2 ------------->Tension Rope 2

Paymemoney said:
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.

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?

Paymemoney said:
How do you know which glider is which?
They start counting from the plane.

oh ic, i understand now thanks for the help.

Can anyone answer my other question.

Paymemoney said:
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?

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.)

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?

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.

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.

Paymemoney said:
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

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

Paymemoney said:
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!

Paymemoney said:
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.

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.)

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

Paymemoney said:
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.

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.

Paymemoney said:
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).

so for question c) the frictional force would be:
6297-Friction = 2989

Friction = 3308N

Paymemoney said:
so for question c) the frictional force would be:
6297-Friction = 2989

Friction = 3308N
Right!

1. What is motion?

Motion is the change in position of an object over time. It can be described in terms of distance, speed, and direction.

2. What is the difference between speed and velocity?

Speed is a measure of how fast an object is moving, while velocity is a measure of how fast an object is moving in a specific direction.

3. How is acceleration calculated?

Acceleration is calculated by dividing the change in velocity by the change in time. It is expressed in units of distance per time squared (m/s²).

4. What is the difference between uniform and non-uniform motion?

Uniform motion is when an object moves at a constant speed in a straight line, while non-uniform motion is when an object's speed or direction changes over time.

5. How does Newton's first law of motion apply to everyday life?

Newton's first law of motion states that an object will remain at rest or in uniform motion in a straight line unless acted upon by an external force. This can be seen in everyday life, such as when a car stays in motion until brakes are applied or when a book remains at rest until someone picks it up.