Finding Equivalent Force on a Stubborn Mule

In summary: What is the hypotenuse?If you want the force to cancel it, you want to add a vector of the same magnitude but in the opposite direction. So what is the opposite of your resultant?In summary, for problem #1, you need to find the resultant of the two given forces by adding their x and y components, and then using trigonometry to find the magnitude and direction of the resultant. To find the force that cancels the resultant, you need to add a vector of the same magnitude but opposite direction to the resultant.
  • #1
willingtolearn
62
0

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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  • #2
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?
 
  • #3
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...
 
  • #4
Problem 1 fixed
 
  • #5
willingtolearn said:
Problem 1 fixed

cool. now you should be able to get the components, as Mindscrape mentioned.
 
  • #6
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.
 
  • #7
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.
 
  • #8
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...
 
  • #9
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?
 

1. How do you determine the equivalent force on a stubborn mule?

The equivalent force on a stubborn mule can be determined by measuring the amount of force needed to counteract the mule's resistance. This can be done by using a force gauge or by observing the mule's behavior and estimating the force required.

2. What factors can affect the equivalent force on a stubborn mule?

The equivalent force on a stubborn mule can be affected by various factors such as the mule's weight, the terrain it is on, the amount of resistance it is putting up, and the person or object applying the force.

3. Can the equivalent force on a stubborn mule be calculated mathematically?

Yes, the equivalent force on a stubborn mule can be calculated using Newton's Second Law of Motion, which states that force equals mass times acceleration. By measuring the mule's mass and observing its acceleration, the equivalent force can be calculated.

4. How can the equivalent force on a stubborn mule be reduced?

The equivalent force on a stubborn mule can be reduced by using techniques such as positive reinforcement, proper training, and understanding the mule's behavior and needs. By creating a positive and comfortable environment, the mule may be less resistant and require less force to be moved.

5. Is it possible to completely eliminate the need for an equivalent force on a stubborn mule?

It is unlikely to completely eliminate the need for an equivalent force on a stubborn mule, as resistance and stubbornness are natural behaviors for many animals. However, by using effective training methods and creating a positive relationship with the mule, the required force may be reduced.

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