Work problem: Pulling wagon and finding angle

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Homework Help Overview

The problem involves calculating the force applied by a parent pulling a wagon and determining the angle of that force. The context includes a scenario with a combined mass of 50kg for the wagon and child, work done of 2.2 x 10^3 J over a distance of 60m, and a coefficient of friction of 0.26.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between work, force, and angle, questioning whether the angle is necessary for calculating the applied force. There is discussion about the implications of constant speed and the role of friction in the problem.

Discussion Status

Participants are actively engaging with the problem, raising questions about the correctness of their assumptions and calculations. Some have provided alternative interpretations of the normal force and its relationship to the pulling force and angle. There is no explicit consensus, but multiple lines of reasoning are being explored.

Contextual Notes

Participants are grappling with the definitions and relationships between the forces involved, particularly regarding the normal force and its dependence on the angle of the applied force. There are indications of confusion around the arithmetic and the application of trigonometric identities.

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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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