- #1

Deceit

- 8

- 0

## 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!