Work problem: Pulling wagon and finding angle

AI Thread Summary
The discussion revolves around solving a physics problem involving a parent pulling a wagon with a combined mass of 50kg. The work done by the parent is 2.2 x 10^3 J over a distance of 60m, with a coefficient of friction of 0.26. Participants are attempting to determine the magnitude of the force applied by the parent and the angle of that force. The calculations involve using the work-energy principle and friction equations, leading to a derived angle of approximately 73 degrees and a force of 125.41N. The conversation highlights the complexities of relating force, angle, and friction in the context of constant speed motion.
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Homework Statement


In many neighborhoods, you might see parents pulling youngsters in a four-wheeled 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

1) Determine the magnitude of the force applied by the parent

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

The Attempt at a Solution



1)
Given:
m = 50kg
W = 2.2 x 10^3 J
d = 60m
mu(kinetic) = 0.26

W = Fa cos(angle) d
Fa = W / cos (angle) d

Do I need to find the angle to solve this?

2)

Constant speed so no acceleration, force applied = kinetic friction
Fk = (Fn)(mu kinetic)
(Fn)(mu kinetic) = W / cos(angle)d
No vertical movement so Fn = weight = ma
cos(angle) = W / (m)(a)(mu kinetic)(d)

cos(angle) = (2.2 x 10^3 N/m) / (50kg)(9.8N/kg)(0.26)(60m)
(angle) = 73 degrees

Go back to 1):

Fa = W / cos (angle) d
Fa = (2.2 x 10^3 N/m) / cos(73)(60m)
Fa = 125.41N

Is this correct? Or does that value of work already have the angle included in its calculation..?
 
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No vertical movement so Fn = weight = ma
This is not true. If F is the pulling force on the wagon, F*sin(theta) will act in the opposite direction to mg. So net Fn will be mg - Fsin(theta).
 
Ok this is what I have so far:
W = Fcos(angle)d
Fa = W / cos(angle)d
Constant speed so Fa = Fk
Fk = W / cos(angle)d
Fk = Fn(mu kinetic) so
Fn(mu kinetic) = W / cos(angle)d
cos(angle) = W / Fn(mu kinetic)
But Fn = mg[sin(angle)]. Now I am stuck with the arithmetic: (assuming I did everything else right?)
cos(angle) sin(angle) = W / mg (mu kinetic)
Do they turn into tan somehow?
 
Fn = mg[sin(angle)].
This is not correct.
Fn = mg - Fa*sin(theta).
 
Fn(mu kinetic) = W / cos(theta)d
Fn = mg - Fa*sin(theta) so,
[mg - Fa*sin(theta)] (mu kinetic) = W / cos*(theta) d
[mg - Fa*sin(theta)][cos*(theta)] = W / (mu kinetic) d
Not sure what to do now :/
Can I factor our (theta)? Does Fa = ma?
 
Find Fa*cos(theta)
Rewrite {[mg - Fa*sin(theta)][cos*(theta)] = W / (mu kinetic) d}
as cos(theta) =...
Square both the sides and write [1-sin^2(theta)] =...
Then solve the quadratic to find the sin(theta) and then Fa.
 
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