Finding Equivalent Force on a Stubborn Mule

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Homework Help Overview

The discussion revolves around two physics problems involving forces and motion. The first problem concerns finding an equivalent single force from two forces acting on a stubborn mule, while the second problem involves calculating the velocity of rain with respect to a moving car. Both problems require an understanding of vector components and resultant forces.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the need to find components of forces and question the orientation of diagrams. There are attempts to clarify what is meant by equivalent forces and resultant forces. Some participants express confusion about the second problem and seek further clarification on the concepts involved.

Discussion Status

The discussion is ongoing, with participants providing hints and clarifications about the problems. Some have offered guidance on how to approach the first problem, while others are still grappling with the second problem. There is a mix of understanding and confusion among participants, indicating a productive exploration of the topics.

Contextual Notes

Participants are working under the constraints of homework rules, which may limit the amount of direct assistance they can receive. There are indications of differing interpretations of the problems, particularly regarding the setup and calculations involved.

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Homework Statement


#2
Two people pull on a stubborn mule, as seen from a helicopter in Figure 1. Find (a) the single force that is equivalent to the two forces shown, and (b) the force that a third person ould have to exert on the mule to make the net force equal to zero.

Figure 1:
http://img526.imageshack.us/img526/8856/123456gn4nr0.png
the blue on the left and right, represent for a person.#3
A car travels due east with a speed of 50km/h. Rain is falling vertically with respect to the Earth. The traces of the rain on the side windows of the car make a angle of 60* with the vertical. Find the velocity of the rain with respect (a) the car and (b) the Earth.
2. The attempt at a solution
I am totally got no idea with the problem #1. Can you guys give me some hints or help me please?

#2
Is this right right picture for the problem ?
http://img523.imageshack.us/img523/6784/123456rq9.png
 
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Problem number 1: hint/method = find the components

Problem number 2: it is right geometrically, but why do you have it oriented all strange like?
 
For the first problem, the angles don't look right. The blue angle is the smaller angle in the picture... but it is 60, while the bigger angle is 45...

For the second problem... Yes, your picture is correct. If you are unsure though, I recommend writing the vectors in i/j form... [tex]\vec{i}[/tex] for east west... [tex]\vec{j}[/tex] for the vertical direction... take r as the speed of the rain... so what is the velocity of the car with respect to the earth... what is the velocity of the rain with respect to the Earth (in terms of r)... i,j are just unit vectors...
 
Problem 1 fixed
 
willingtolearn said:
Problem 1 fixed

cool. now you should be able to get the components, as Mindscrape mentioned.
 
What they mean by Find (a) the single force that is equivalent to the two forces shown, and (b) the force that a third person ould have to exert on the mule to make the net force equal to zero.
 
They mean, if you could express the two forces at those two angles, as just one force at one angle, what would that force be? Then they want to know what force cancels it.
 
willingtolearn said:
What they mean by Find (a) the single force that is equivalent to the two forces shown

the resultant of the two forces. The sum...
 
willingtolearn said:
What they mean by Find (a) the single force that is equivalent to the two forces shown, and (b) the force that a third person ould have to exert on the mule to make the net force equal to zero.

They mean to find the resultant of the forces.In problem 1. While in problem 2 they are asking you to find a force which is equivalent to the resultant of the other two forces in magnitude but opp in direction. So for all the forces the following should hold.

[tex]\sum{F_x}[/tex]=0

[tex]\sum{F_y}[/tex]=0
 
  • #10
learningphysics said:
the resultant of the two forces. The sum...

Got it !
Mindscrape said:
They mean, if you could express the two forces at those two angles, as just one force at one angle, what would that force be? Then they want to know what force cancels it.
your explanation is a little complicated !

FedEx said:
They mean to find the resultant of the forces.In problem 1. While in problem 2 they are asking you to find a force which is equivalent to the resultant of the other two forces in magnitude but opp in direction. So for all the forces the following should hold.
Got the first one, but totally out of the second one ! It too complicated to me!


Thanks you all ! You guys have a good day !
 
  • #11
Sorry if it is complicated, but you need to realize it's all relative. If the mule were in a black box how would he know whether it was two people pushing on him, 8 people pushing on him, or just one person pushing on him? The particle, mule in your problem, doesn't care how many forces are acting on it, it just cares about the sum of the forces, and how they are superposed onto it.
 
  • #12
willingtolearn said:
Got the first one, but totally out of the second one ! It too complicated to me!


Thanks you all ! You guys have a good day !

You've got the resultant of the two forces... now you need a force out the original two forces (which is exactly the same as the force to cancel out the resultant of the first two forces)...

If you have a vector... and you want another vector to add to it, to make the sum zero... then you add a vector of the same magnitude but opposite direction...

so part b), is just the same as part a), except the opposite direction.
 
  • #13
I thought I got it, but you confused me !
#1
a) after i found the component of first force and second force, then just add the y component to get the resultant ! Is that right
b) still confuse
 
  • #14
No, so for the first one you need to add all the x components and y components:

[tex]F_{net_x} = F_1_x + F_2_x[/tex]

and

[tex]F_{net_y} = F_1_y + F_2_y[/tex]

Then the net resultant is

[tex]||F_{net}|| = \sqrt{F_{net_x}^2 + F_{net_y}^2}[/tex]

What would the direction be?

How would you find the force that cancels F_net?

What are you confused about on the second one?
 
  • #15
F_net = 198 N
I only get that
 
  • #16
What do you mean you only get that? I don't know if that is the only part you understand, or the only part you have gotten because it's the only one I explicitly told you how to do. If you want the direction, think about trig and how to find angles of a triangles. What do your x and y vectors look like with your resultant vector, a bit like a triangle?
 

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