The Vector nature of Newton's Second Law

[Mathcaptor]

Homework Statement

A wagon carries a child. Together mass is 28.5kg. You pull on the handle of the wagon at an angle of 40degrees from the horizontal. The wagon travels over a level horizontal sidewalk. A force of friction of 12.0N acts on the wagon.

a) What is the net force acting on the wagon?

Homework Equations

Fnet = F1 + F2
Trigonometry

The Attempt at a Solution

I missed a day of Physics class but the teacher gave me this assignment. I thought it was suppose to be solving right triangles but now, I'm not too sure. I've been looking through my notes! Please, I need a visual representation.

 

[rock.freak667]

Draw the wagon, the frictional force acts opposite to the motion. Resolve the applied force into vertical and horizontal components.

Can you do those two?

 

[Mathcaptor]

Draw the wagon, the frictional force acts opposite to the motion. Resolve the applied force into vertical and horizontal components.

Can you do those two?

Would it look like this?

 

[rock.freak667]

Yes, it would, though on the handle, there should be a force going away from the wagon since you are pulling it.

 

[Mathcaptor]

Yes, it would, though on the handle, there should be a force going away from the wagon since you are pulling it.

So there should be an arrow on the handle line. With that, which vector's magnitude am I suppose to figure out...? There are unknown sides to that triangle.

 

[rock.freak667]

So there should be an arrow on the handle line. With that, which vector's magnitude am I suppose to figure out...? There are unknown sides to that triangle.

Put the unknown force as, F. Draw the right angled trianle now.
Can you express the sides of right angled triangle in terms of F?

 

[Mathcaptor]

Put the unknown force as, F. Draw the right angled trianle now.
Can you express the sides of right angled triangle in terms of F?

Would it look like that? and I'm suppose to find Fx, Fy and F?

I'm trying to figure out the net force and that includes friction I believe.

 

[rock.freak667]

Would it look like that? and I'm suppose to find Fx, Fy and F?

I'm trying to figure out the net force and that includes friction I believe.

yes, that is how it would look.

Now,
[tex]sin\theta=\frac{opposite}{hypotenuse}[/tex]

then

[tex]sin40 = \frac{F_y}{F}[/tex]

Correct? Then F_y=?

Now consider what cos40 should be

 

[Mathcaptor]

yes, that is how it would look.

Now,
[tex]sin\theta=\frac{opposite}{hypotenuse}[/tex]

then

[tex]sin40 = \frac{F_y}{F}[/tex]

Correct? Then F_y=?

Now consider what cos40 should be

I understand this but the only thing is that F, F_y and F_x have no values.

 

[rock.freak667]

I understand this but the only thing is that F, F_y and F_x have no values.

Forget that they have no values for the moment. (After you resolve the forces, you will be able to get the value for F)

[tex]sin40 = \frac{F_y}{F} \Rightarrow F_y= ?[/tex]

 

[Mathcaptor]

Forget that they have no values for the moment. (After you resolve the forces, you will be able to get the value for F)

[tex]sin40 = \frac{F_y}{F} \Rightarrow F_y= ?[/tex]

F_y/Sin40 = F

 

[rock.freak667]

F_y/Sin40 = F

Other way around, F_y=Fsin40

Similarly cos40= Fx/F, so Fx=Fcos40 right?

In your Free body diagram, include the weight of the wagon.

EDIT: Now that you have Fx and Fy found. Knowing that the weight acts downwards and Fy upwards. Is there any net motion upwards?

 

[Mathcaptor]

Other way around, F_y=Fsin40

Similarly cos40= Fx/F, so Fx=Fcos40 right?

In your Free body diagram, include the weight of the wagon.

Alright, I understand the equations, but how am I suppose to find the values now?
What do you mean by include the weight of the wagon...? Is it another vector?

 

[rock.freak667]

Alright, I understand the equations, but how am I suppose to find the values now?
What do you mean by include the weight of the wagon...? Is it another vector?

Well the question stated that the mass of the wagon and child was 28.5kg, so it has an assoiciated weight. Yes, it is another vector.

 

[Mathcaptor]

Well the question stated that the mass of the wagon and child was 28.5kg, so it has an assoiciated weight. Yes, it is another vector.

Is it possible for yourself to draw it because I'm having trouble visualizing it... If I add a weight vector, wouldn't it throw me off with the fx fy and f vectors...

 

[rock.freak667]

http://img352.imageshack.us/img352/6447/diagramii8.jpg

So, when you pull a wagon, it only moves horizontal right? Meaning that there is no vertical movement, therefore, the forces Fy and weight should balance.

So Fy=Weight

=> Fsin40=28.5g where g is the acceleration due to gravity. Can you now find F?

 

[Mathcaptor]

http://img352.imageshack.us/img352/6447/diagramii8.jpg

So, when you pull a wagon, it only moves horizontal right? Meaning that there is no vertical movement, therefore, the forces Fy and weight should balance.

So Fy=Weight

=> Fsin40=28.5g where g is the acceleration due to gravity. Can you now find F?

F = 28.5g / sin40

is that the net force? or do I have to figure out Fx and add F + Fx + Friction(with direction) to find F net

 

[Stovebolt]

I disagree with the statement that Fy = 28.5g. If this was the case, either there would be no friction (as the wagon isn't exerting any force on the ground) or there is a moment acting on the wagon, in which case it will have a rotational acceleration - and that seems much more complicated than this problem is supposed to be. I would say that there should be a net force between the wagon and the ground, meaning you will need one more force in your diagram.

The problem, as originally stated, is not determinate. If the wagon is not moving at a constant velocity, then it is being accelerated (or decelerated). To find the net force, you will need to be given either the acceleration or information on how the frictional force is calculated (are the wheels dragging? or is the friction in the hub of the wheels? In either case, you would need a coefficient for friction). Again, I think this is more complicated than the problem is supposed to be.

So after all that, I believe it is safe to assume the wagon is supposed to be moving at a constant velocity. If this is indeed the case, the problem has just become amazingly easy (check Newton's First Law of Motion). However, I also think the question is actually asking for the force applied on the handle, F. To find this, consider what the sum of the forces in the horizontal direction will be. Once that is determined, you will be able to find Fx, which will then be a simple calculation to find F.

 

[rock.freak667]

Resultant Vertical force=0
=> Fsin40=28.5g...F can be found

Resultant horizontal force,F_net=Fcos40-12...when F is found from above, the resultant horizontal force can be found.

 

[Stovebolt]

The sum of the vertical forces will be equal to zero, but there are three forces acting in the vertical direction:
F(sin 40) (upward force)
28.5g (downward force)

and the normal force of the ground upwards on the wagon (I'll call it F_n)

So
F(sin 40) - 28.5g + F_n = 0

If there was not a force between the wagon and the ground, there would be no frictional forces involved in this problem (no force between the wheels and the ground means the wheels would neither turn nor drag).

So we are left with a missing piece of information - either we need the friction coefficient so we can find F_n, or we need to know what the acceleration of the wagon is.