How Do You Calculate the Acceleration of a Toboggan with Given Forces?

[anna sung]

Homework Statement

A toboggan with a mass of 15kg is being pulled with an applied force of 45 N at an angle of 40degrees to the horizontal. What is the acclecration if the force of friction opposing the motion is 28 N?

Homework Equations

Is this question related to the x and y component rules?
I tried finding the x and y of 45 N and 28N. I don't understand how my answer is not even close to the answer in the book.

The Attempt at a Solution

45N (E40degreeN)
x-component:
45(cos40)=34.37N
y-component:
45(sin40)=28.9N

28N (W40degreeS)
x-component:
28(cos40)=-21.4N
28(sin40)=-18N

X-components
34.37-21.4= 13.07N
Y-components
28.9-18=10.92N
\sqrt{13.07^2+10.92^2}
= 17N
Fun=ma
a=Fun/m
a=17N/15kg
a=1.1m/s^2

The answer in the book is 0.4m/s^2
sorry about this mess. its my first time posting a question.

 

[markpace123]

Your error was inpropper diagram, in this kind of physics the 3 basic rules are draw draw draw.


28N <--- ______) ----> 45Cos(40)

this is a representation of the horizontal component forces acting on the toboggan (the sled).

subtracting them you should get approx:

6.47N the right direction of the screen

therefore then applying F=ma

a = F / m

a = 6.47 / 15

a = 0.431 m/s/s QED =D

Hope this helps mate

Homework Statement

A toboggan with a mass of 15kg is being pulled with an applied force of 45 N at an angle of 40degrees to the horizontal. What is the acclecration if the force of friction opposing the motion is 28 N?

Homework Equations

Is this question related to the x and y component rules?
I tried finding the x and y of 45 N and 28N. I don't understand how my answer is not even close to the answer in the book.

The Attempt at a Solution

45N (E40degreeN)
x-component:
45(cos40)=34.37N
y-component:
45(sin40)=28.9N

28N (W40degreeS)
x-component:
28(cos40)=-21.4N
28(sin40)=-18N

X-components
34.37-21.4= 13.07N
Y-components
28.9-18=10.92N
\sqrt{13.07^2+10.92^2}
= 17N
Fun=ma
a=Fun/m
a=17N/15kg
a=1.1m/s^2

The answer in the book is 0.4m/s^2
sorry about this mess. its my first time posting a question.

 

[anna sung]

oh thanks i get it. but I am still not sure why 45cos40
why do we need to get the x component what does this number represent?

 

[markpace123]

Hello again,

well we are considering only the x component because the motion of the sled is in the horizontal plane, the x plane.

this number the 45cos 40 represents the force, the component force present in the x axis. Basic maths; pythagoras theorem you were right to apply it.

The principle behind this is that the hypotenuse squared is the sum of the x component squared and the y component squared. In this case your hypotenuse is the 45N given.

If you adopt vectors from mathematics you will see why later on, but for this question you have understand that the force to the right is not 45N but a component of that the 45cos 40 which is less than the 45N (because some of that is going to the force in the y direction as well).

In simple terms that 45N stems from two forces which are perpendicular to each other; x and y.

Hope this answers your question =)

 

[markpace123]

haha you're welcome, sorry for confusing you ... I tried my best to describe it but if we had some sort of visual aid I'm sure you would have understood it very well.

If you need anything please feel free to email me or send me a private message. I will do my best =)

All the best with your studies, I assure you physics is an endlessly facinating subject.