Work with Friction: Calculating Force & Angle

In summary, working with friction involves understanding how to calculate force and angle in order to overcome the resistance caused by friction. This requires knowledge of the coefficient of friction, which is a measure of how much friction exists between two surfaces. By using equations such as the one developed by Sir Isaac Newton, it is possible to determine the necessary force and angle needed to overcome friction and successfully move an object. Additionally, the angle at which the force is applied can greatly impact the amount of friction experienced, highlighting the importance of considering both force and angle when working with friction.
  • #1
dator
5
0

Homework Statement


In many neighbourhoods, 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 2.2x10^3 J of work pulling the two 60m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.

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

Homework Equations


F_friction = mu*F_normal
a^2=b^2+c^2
Work=Force*(delta)d*cos(theta)

The Attempt at a Solution


Just want to make sure I did this right. And do I have to include the units when i write formulas out? (N*M) etc. It it important too or it doesn't matter? Thanks in advance!

Variables:
Work = 2.2x10^3 J
(delta)d= 60m
mu = 0.26
Force = ?
Theta =?

In Terms of y:
F_normal=-(mg+Force*sin(theta))

In Terms of x:
F_friction = mu*F_normal

Since speed is constant acceleration=0, therefore F_x+F_friction = 0 or (F_x=-F_friction)

Let:
Force*sin(theta) = F_y
Force*cos(theta) = F_x

In terms of x:

Work = Force*(delta)d*cos(theta)
Force*cos(theta) = Work/(delta)d
F_x = 2.2x10^3/60
F_x = 36.77777776

Therefore F_x+F_friction = 0

F_x+mu*F_normal = 0

Sub in numbers.

36.7+(-1)(0.26)[(50)(-9.8)+Force*sin(theta)= 0
36.7+(-0.26)[-490+Force*sin(theta)]= 0
36.7+127.4+(-0.26)(Force*sin(theta) = 0
Force*sin(theta) = -164.1/-0.26
Force*sin(theta) = 631.1538462
F_y = 631.1538462

F_y = 631.1538462
F_x = 36.7

Force = sqrt(F_y^2+F_x^2)
Force = 632.38N

tan(theta) = opp/adj
opp = F_y
adj = F_x

(theta) = tan^-1(F_y/F_x)
(theta) = 86.6
(theta) = 87

a)cant draw it here, but the F_normal has the F_y going parallel with it, and F_fr is opposing the F_x.

b) 632.38N
c)(theta) = 87
 
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  • #2
Hello dator,

Welcome to Physics Forums!

dator said:

Homework Statement


In many neighbourhoods, 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 2.2x10^3 J of work pulling the two 60m at a constant speed. The coefficient of friction for the surfaces in contact is 0.26.

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

Homework Equations


F_friction = mu*F_normal
a^2=b^2+c^2
Work=Force*(delta)d*cos(theta)

The Attempt at a Solution


Just want to make sure I did this right. And do I have to include the units when i write formulas out? (N*M) etc. It it important too or it doesn't matter? Thanks in advance!

Variables:
Work = 2.2x10^3 J
(delta)d= 60m
mu = 0.26
Force = ?
Theta =?

In Terms of y:
F_normal=-(mg+Force*sin(theta))

In Terms of x:
F_friction = mu*F_normal

Since speed is constant acceleration=0, therefore F_x+F_friction = 0 or (F_x=-F_friction)

Let:
Force*sin(theta) = F_y
Force*cos(theta) = F_x

In terms of x:

Work = Force*(delta)d*cos(theta)
Force*cos(theta) = Work/(delta)d
F_x = 2.2x10^3/60
F_x = 36.77777776

Therefore F_x+F_friction = 0

F_x+mu*F_normal = 0

Sub in numbers.

36.7+(-1)(0.26)[(50)(-9.8)+Force*sin(theta)= 0

According to my calculations, something is not right with the very last line above.

I think the root of the error might come from your equation:
F_normal=-(mg+Force*sin(theta))
The vectors are not adding up (or subtracting up) right. Rearrange the above equation to put all the terms on one side of the equation; then set all that to zero. Compare that to your FBD.

In the future, if might be beneficial to work the problem in smaller steps, rather than trying to combine everything into one big monster step. For example, after you calculate the force of friction (which you did early, and that's okay), calculate the normal force. That way you can check if things make sense before moving on to the next step. :wink:
 
  • #3
collinsmark said:
Hello dator,

Welcome to Physics Forums!
Thank you! :)

collinsmark said:
According to my calculations, something is not right with the very last line above.

I think the root of the error might come from your equation:

The vectors are not adding up (or subtracting up) right. Rearrange the above equation to put all the terms on one side of the equation; then set all that to zero. Compare that to your FBD.

Bah, I should plugged the numbers back in and checked the statement I made, F_x+F_friction=0. The F_y I solved for does not give me a true statement.

Statements should be as follows then,
F_normal = -(m*-g+Force*sin(theta))
F_normal = (mg-Force*sin(theta))

Therefore, F_x+F_friction = 0 is:

Force*cos(theta)+(mu)(mg-Force*sin(theta)) = 0
36.7+(0.26)(490-Force*sin(theta)) = 0
36.7+127.4-(0.26)(Force*sin(theta)) = 0
Force*sin(theta) = -164.1/-0.26
Force*sin(theta) = 631.1538462

Double check...

36.7+(0.26)(490-631.1538462)=-0.0000001
So basically zero, blame it to roundoff.

Does that look good now?

Thanks, in advance!
 
  • #4
dator said:
Thank you! :)
dator said:
Thank you! :)
Force*cos(theta)+(mu)(mg-Force*sin(theta)) = 0
36.7+(0.26)(490-Force*sin(theta)) = 0

No, something is still not quite right. :wink: Both the frictional force, and the horizontal pulling force are both in the same direction. They need to be in opposite directions. Try placing a negative sign in front of the 36.7.


Double check...

36.7+(0.26)(490-631.1538462)=-0.0000001
So basically zero, blame it to roundoff.

Does that look good now?

Thanks, in advance!

Your double checking should show that something is not right. The vertical pulling of the parents has a greater magnitude than the gravitational force. The parents would pulling the wagon so hard that it would accelerate up into the air! :-p
 
  • #5
degrees

I would say that your solution looks correct based on the information given. It is always important to include units when writing out formulas, as they provide important context and ensure that the calculations are done correctly. In this case, the units for force would be Newtons (N) and the units for work would be Joules (J). The angle theta would be in degrees (°). It is also important to note that the angle at which the parent is pulling the wagon would be measured from the horizontal, not the vertical, so the angle would be 90-87 = 3 degrees. Overall, your solution appears to be well thought out and accurate.
 

Related to Work with Friction: Calculating Force & Angle

What is friction?

Friction is a force that resists the motion of two objects that are in contact with each other. It is caused by the irregularities in the surfaces of the objects and can be affected by factors such as the type of material, surface area, and force applied.

Why is it important to calculate force and angle in work involving friction?

Calculating force and angle in work involving friction is important because it allows us to understand and predict the behavior of objects in contact with each other. This information is crucial in many fields, such as engineering and physics, as it helps in designing and building structures and machines that can withstand the effects of friction.

How do you calculate the force of friction?

The force of friction can be calculated using the formula F = μN, where F is the force of friction, μ is the coefficient of friction, and N is the normal force. The coefficient of friction is a constant value that depends on the type of materials in contact, while the normal force is the force perpendicular to the surface of contact.

What is the angle of friction?

The angle of friction is the angle at which an object on a surface will begin to slide due to the force of friction. It is also known as the angle of repose and is equal to the arctan of the coefficient of friction. It is an important factor to consider in determining the stability of objects on an inclined surface.

How does friction affect the motion of objects?

Friction can either increase or decrease the motion of objects, depending on the direction and magnitude of the applied force. In general, friction causes objects to slow down and eventually come to a stop. However, in some cases, friction can also be used to increase the motion of objects, such as in the case of car brakes where friction is used to slow down the vehicle.

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