Work problem: Pulling wagon and finding angle

In summary, the conversation discusses a situation where a parent is pulling a child and a wagon with a combined mass of 50kg at a constant speed of 60m. The parent does 2.2 x 10^3 J of work and the coefficient of friction is 0.26. The conversation then explores finding the magnitude of the force applied by the parent and the angle at which the force is applied. The solution involves finding the force of kinetic friction and using trigonometric functions to find the angle and the force applied.
  • #1
zeion
466
1

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..?
 
Physics news on Phys.org
  • #2
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).
 
  • #3
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?
 
  • #4
Fn = mg[sin(angle)].
This is not correct.
Fn = mg - Fa*sin(theta).
 
  • #5
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?
 
  • #6
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.
 

1. What is the work problem involving pulling a wagon and finding the angle?

The work problem involves calculating the amount of work required to pull a wagon at a certain angle, given the amount of force applied and the distance traveled.

2. How is the angle related to the work required?

The angle is directly related to the amount of work required. As the angle increases, the amount of work required also increases.

3. What factors affect the amount of work required to pull the wagon?

The amount of work required is affected by the angle of pull, the weight of the wagon, and the distance traveled. It is also influenced by external factors such as friction and air resistance.

4. How can the work problem be solved?

The work problem can be solved by using the formula W = Fdcosθ, where W is work, F is force, d is distance, and θ is the angle. The angle can be determined using trigonometric functions.

5. Why is it important to understand work problems involving pulling a wagon and finding the angle?

Understanding work problems is important for many practical applications, such as determining the amount of force needed to move an object at a certain angle, or calculating the efficiency of machines. It also helps in understanding the principles of physics and the relationship between force, work, and energy.

Similar threads

  • Introductory Physics Homework Help
Replies
10
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
689
  • Introductory Physics Homework Help
Replies
1
Views
1K
  • Introductory Physics Homework Help
Replies
10
Views
313
  • Introductory Physics Homework Help
Replies
9
Views
1K
  • Introductory Physics Homework Help
Replies
1
Views
340
  • Introductory Physics Homework Help
Replies
2
Views
208
  • Introductory Physics Homework Help
Replies
3
Views
4K
  • Introductory Physics Homework Help
Replies
2
Views
2K
  • Introductory Physics Homework Help
2
Replies
38
Views
2K
Back
Top