Gr. 12 Energy Problem - Pulling a Wagon

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Homework Statement



This is from a book for a correspondence course I'm taking. I don't have access to teachers, so hopefully some of you can fill that role :)

A parent is pulling a wagon. The child and the wagon have a combined mass of 50kg and the adult does 2.2 x 10^3 J of work pulling the two 60m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.

a) Determine the magnitude of the force applied by the parent.
b) Determine the angle at which the parent is applying this force.


Homework Equations



W = F * d
F(k) = u(k) * F(N)


The Attempt at a Solution



For the horizontal component of F(A):
W = F * d
2.2 x 10^3 J = F(Ah) * (60m)
F(Ah) = 37 N

To find the vertical component of F(A):
F(v) = F(N) + F(g) + F(Av)
(v means vertical)

F(N) = F(k) / u(k)
and, because the horizontal forces balance, F(Ah) = F(k)
Because the forces are balanced, F(v) = 0 so:

0 = (F(f) / u(f)) + (m * g) + F(Av)
-F(Av) = (37N / 0.26) + (50kg)(-9.8N/kg)
-F(Av) = -350N
F(Av) = 350N

Plugging in the component forces into the pythagorean theorem, I find F(A) = 350N
I'm only using two sig digs because they only use one in the question, and the book instructs to use one more sig dig during intermediate calculations than the original question uses. So, for the final answer, F(A) should actually be 400N to one sig dig.

As for the angle, using trig I find it to be 90 degrees.

Now, this angle is very steep (obviously), which is leading me to believe I did something wrong. Does this solution make sense or am I thinking about this the wrong way? Any insight is appreciated!
 

Answers and Replies

  • #2
ehild
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Be careful with the significant digits. It is better to use more than required than to loose accuracy. The instructions says that you have to use one digit more than the number of significant digits in the original data. That is two in this problem so do the calculations with 3 digits, and round off to two at the end.

You calculated the total applied force with two digits, which happened to be the same as the vertical component with that accuracy. You lost one digit and got the angle with one significant digit: 90 degree, which is not a sensible answer. And it was unnecessary to calculate the magnitude of the applied force. Calculate the angle from the tangent: you get 84 degrees. That is acceptable, is not it?

ehild
 
  • #3
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Be careful with the significant digits. It is better to use more than required than to loose accuracy. The instructions says that you have to use one digit more than the number of significant digits in the original data. That is two in this problem so do the calculations with 3 digits, and round off to two at the end.

You calculated the total applied force with two digits, which happened to be the same as the vertical component with that accuracy. You lost one digit and got the angle with one significant digit: 90 degree, which is not a sensible answer. And it was unnecessary to calculate the magnitude of the applied force. Calculate the angle from the tangent: you get 84 degrees. That is acceptable, is not it?

ehild

The first part of the question did ask for the magnitude of the applied force. As for the significant digits, the question gave values of 50kg and 60m, each of which would technically only have one significant digit, no?

Thanks for the reply
 
  • #4
ehild
Homework Helper
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The other data clearly have 2 significant digits. 0 can be a significant digit, why not? So 50kg and 60 m have also two significant digits. If they wanted to use only one significant digit, they would have written them as 5 x 10^1 and 6 x 10^1.

If you see that the value of length, for example is 300000 m, it is given with 6 significant digits. The same would be 3.0x10^5 m with two significant digits.

OK, the magnitude of the force was also asked. But use three digits in the calculation. And determine the angle from the ratio of the vertical and horizontal components.


ehild
 
  • #5
8
0
The other data clearly have 2 significant digits. 0 can be a significant digit, why not? So 50kg and 60 m have also two significant digits. If they wanted to use only one significant digit, they would have written them as 5 x 10^1 and 6 x 10^1.

If you see that the value of length, for example is 300000 m, it is given with 6 significant digits. The same would be 3.0x10^5 m with two significant digits.

OK, the magnitude of the force was also asked. But use three digits in the calculation. And determine the angle from the ratio of the vertical and horizontal components.


ehild

According to the book, any trailing zeros without a decimal place are not considered significant. But, I guess it really doesn't make sense to only use one sig dig in this case, so I'll consider the zeros significant.

Thanks for the help
 

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