Is it possible to pull a wagon without wheels?

[benca]

Homework Statement

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

Determine the magnitude of the force applied by the parent.

Determine the angle at which the parent is applying this force.

Relevant Equations

W = f*cos*d
Ff = μK*Ff

Fa = Force applied on the cart by the parent
Fax = x component For the force applied on the cart by the parent
Fay = y component For the force applied on the cart by the parent
Ff = force of friction

The method I thought to use was to solve for Facos, Fasin and then use pythagorean's theorem to solve for Fa

W = Facosd
Facos= W/d
= 2.2x10^3 J / 60m
=36.7 N

Fax= Ff
Fax=μK*Fn
Fn = Fax/μK
= 36.7 N / 0.26
= 141.2 N

Fn + Fay = Fg
Fay = Fg - Fn
= (50kg*-9.8m/s^2) - 141.2 N
= -632 N

At this point I'm almost certain this is wrong, mostly because Fay came out as a negative when it is acting in the upward (positive) direction. and even if it was positive, 631 N seems too high.

One of the main problems I have when doing these questions is that I don't really understand when to put a positive or negative sign. Fax = Ff for example. It seems like the magnitude of these two forces should be the same, but should I write -Ff instead?

 

[PeroK]

You have to decide what direction (up or down) is positive. If you take $g = -9.81m/s^2$, then up is positive. Have you been consistent in that?

 

[benca]

I went over it again and I think I have been consistent, at least in the y direction. Set right/up to positive and left/down to negative. The work given is positive so that means it's going in the same direction as the motion of the cart (to the right)

Fn + Fay = Fg

The numbers I substituted for Fn was positive and negative for Fg.

I'm unsure of Fax = Ff. Although looking at it again, I think Ff should be negative because I'm not substituting a negative number, but expanding it and solving for one of the variables. Does that make sense or am I missing something?

Adding a negative sign would produce Fay = - 349 N. Still negative.

I'm still having trouble seeing where I went wrong with this: Fn + Fay = Fg

 

[PeroK]

I went over it again and I think I have been consistent, at least in the y direction. Set right/up to positive and left/down to negative. The work given is positive so that means it's going in the same direction as the motion of the cart (to the right)

Fn + Fay = Fg

The numbers I substituted for Fn was positive and negative for Fg.

I'm unsure of Fax = Ff. Although looking at it again, I think Ff should be negative because I'm not substituting a negative number, but expanding it and solving for one of the variables. Does that make sense or am I missing something?

Adding a negative sign would produce Fay = - 349 N. Still negative.

I'm still having trouble seeing where I went wrong with this: Fn + Fay = Fg

The simple approach is that without $F_y$ the downward normal force has a magnitude of $mg$.

If the actual normal force is less than this then there must be an upward force of magnitude $F_y = mg - F_n$.

Where all quantities are positive.

You can check that in the cases where $F_n = 0$: the upward force must equal the gravitational force. And where $F_n = mg$, where there is zero upward force.

This saves you getting into a tangle with what's positive and negative.

 

[PeroK]

PS note that the are of course two normal forces, one downwards and one upwards of equal magnitude. You need to decide which of these forces you are talking about in your equations.

 

[PeroK]

I'm still having trouble seeing where I went wrong with this: Fn + Fay = Fg

The vertical forces on the object must be balanced, so we have:

$F_n + F_y + F_g = 0$

Where $F_n$ in this case is the force of the ground on the object, which is upwards, hence positive.

$F_g$ on the other hand is downwards, hence negative.

This should result in a positive value for $F_y$, I.e. upwards.

 

[benca]

PS note that the are of course two normal forces, one downwards and one upwards of equal magnitude. You need to decide which of these forces you are talking about in your equations.

Alright, I'm still wrapping my head around the idea of two normal forces and the difference between a downwards normal force and the force of gravity. I think I need to go over some material again.

The vertical forces on the object must be balanced, so we have:

Fn+Fy+Fg=0

This actually does make a lot of sense, thanks for clearing that up.

So,
Fn+Fy+Fg=0
Fy=-Fn-Fg
Fy=-141N-50kg(-9.8m/s^2)
Fy=349N

√349N^2+36.7N^2=F
351N=F

However, using trig, this gives and angle of 84° The numbers seem odd for this scenario.

W = Fcosd
Fcos=W/d
Fcos=2.2x10^3 J / 60m
Fcos=36.7 N

Is this an accurate calculation of Fx?

 

[benca]

Also, if there is no acceleration in the x direction, does that mean Fx+Ff=0 ?

 

[PeroK]

Also, if there is no acceleration in the x direction, does that mean Fx+Ff=0 ?

Yes. I think you did use that!

And the numbers look odd in this problem too. I think the large angle you got is correct. The wagon would be tipped right up if you analyzed further, but I guess the question setter just picked some numbers out of the air.

If you think about it, the man is nearly lifting the 50kg off the ground!

 

[benca]

And the numbers look odd in this problem too. I think the large angle is correct. The wagon would be tipped right up if you analyzed further, but I guess the question setter just picked some numbers out of the air.

oh my, I think I'm going to have to get used to odd numbers in this course. Thanks

 

[haruspex]

the difference between a downwards normal force and the force of gravity

A common error is to confuse direct forces with indirect forces.
The direct force experienced by the ground is the downward normal force from the wagon. The ground does not "know" the causes of this. The wagon directly experiences an equal and opposite (so upward) normal force from the ground, the force of gravity and the applied force from the adult.

Wrt the calculation, a very good habit to get into is keeping everything algebraic as long as possible, only plugging in numbers at the end. It has many benefits, including, often, greater precision.
E.g. in this case you could have obtained $\tan(\theta)=\frac{mgs}W-1$, where s is the distance pulled, etc.

Btw, seems this wagon has no wheels!