Final Exam Q6: Finding Problem with Torque Equation

In summary: No, you're not making it harder. You're just doing it in a different way. Instead of figuring out the equation for the triangle, you can just use the Pythagorean Theorem. Here's one way to look at it. To get across as fast as possible, just swim straight across (with respect to the water, which will drag you downstream, but who cares?). How long does it take the water to drag the swimmer the given distance downstream? Since he swims across the river in that same time, what must be his swimming speed (with respect to the water)?It takes the water 10 seconds to drag the swimmer the given distance downstream. So the
  • #71
Hootenanny said:
The co-efficent of friction is given by;

[tex]\mu = \frac{F}{mg}[/tex]

Where F is the maximum frictional force. You have to take into account the force that the tractor is pulling.

~H
um so what i did was add all the weights multiply by 9.8 is that even right and then i divided by the tractor weight or am i missing something
 
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  • #72
Alt+F4 said:
um so what i did was add all the weights multiply by 9.8 is that even right and then i divided by the tractor weight or am i missing something

No, I'm afraid it isn't right. The reaction force (mg) is due to the tractor's weight only. The F is the force required to accelerate the total mass at 1.4 m.s-2. Do you follow?

~H
 
  • #73
Hootenanny said:
No, I'm afraid it isn't right. The reaction force (mg) is due to the tractor's weight only. The F is the force required to accelerate the total mass at 1.4 m.s-2. Do you follow?

~H
o yes, thanks
 
  • #74
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp03

Question 23

I did Centripital Acceleration * weight

50 * .5 = 25 is this the right way?
 
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  • #75
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp03

Question 23

I did Centripital Acceleration * weight

50 * .5 = 25 is this the right way?
That's fine. (I assume you meant to write mass, not weight.)
 
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  • #76
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05 Last question, i still don't get how to set it up incase he asks the same question and wants the resultant velocity

so

(1.5*10^5)(8) = (-4*1.5*10^5) + (1.5*10^6*X)

Is this the setup
 
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  • #77
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05 Last question, i still don't get how to set it up incase he asks the same question and wants the resultant velocity

so

(1.5*10^5)(8) = (-4*1.5*10^5) + (1.5*10^6*X)

Is this the setup
That's how I'd do it.
 
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  • #78
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp05

Question 6

So what i was thinkin is to do it this way


so work done by friction = W = *.43*9.8*12 = 50.568

Total Kinetic Energy = MGH + .5 MV^2 = (9.8)(9) + .5 (20^2) = 288.2

288.2-50.568 = 237.632

Edit: well it works if i don't take in effect the PE of the hill but why not?
 
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  • #79
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp05

Question 6

So what i was thinkin is to do it this way


so work done by friction = W = *.43*9.8*12 = 50.568

Total Kinetic Energy = MGH + .5 MV^2 = (9.8)(9) + .5 (20^2) = 288.2

288.2-50.568 = 237.632

Edit: well it works if i don't take in effect the PE of the hill but why not?
Why should the 9 m hill matter? Since there's no friction, the KE is the same before and after climbing that hill.
 
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  • #80
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
 
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  • #81
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 16

So i know that Change in kinetic = Uk g * D

So what i did was (22)(9.8) + (9)(9.8) = X * 9.8 *53

wat am i doing wrong
 
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  • #82
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 16

So i know that Change in kinetic = Uk g * D

So what i did was (22)(9.8) + (9)(9.8) = X * 9.8 *53

Your problem is this equation. Why have you added additional GPE [+ (9)(9.8)]?

~H
 
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  • #83
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 19

So Wnet = 8.6483
Wnet - Force of friction - Force of gravity
I did get the above questions right so i know the numbers are right


8.6483 = .5 mV^2

V = 2.9773


The 3 is the ending velocity
(-2)(2.9773) - (2)(3) = -11.9546 ( Total Momemtum)

Fave T = P

X (.06) = (11.9546)

X = 200

Is this how it is done
 
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  • #84
WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?
 
  • #85
Alt+F4 said:
WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?

In what context? What is the full question?

~H
 
  • #86
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa04

Question 19

So Wnet = 8.6483
Wnet - Force of friction - Force of gravity
I did get the above questions right so i know the numbers are right


8.6483 = .5 mV^2

V = 2.9773


The 3 is the ending velocity
(-2)(2.9773) - (2)(3) = -11.9546 ( Total Momemtum)

Fave T = P

X (.06) = (11.9546)

X = 200

Is this how it is done

Why have you gone through the trouble of calculating the velocity just before impact when it is given in the question? :wink: But yes you working is correct. The change in momentum will be -6m and you were correct to use the impulse relationship.

~H
 
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  • #87
Suppose the mass of the string in the above problem is M0. What is the mass of a new string (in terms of M0 ), if the fundamental frequency of the string is doubled compared to the previous value, but the length and tension in the string are the same as before?

Now does this Mean Frequncy * 2 or frequency / 2
 
  • #88
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
let me just quote this so it doesn't get lost
 
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  • #89
Alt+F4 said:
Suppose the mass of the string in the above problem is M0. What is the mass of a new string (in terms of M0 ), if the fundamental frequency of the string is doubled compared to the previous value, but the length and tension in the string are the same as before?

Now does this Mean Frequncy * 2 or frequency / 2

It means the fundemetal fequency is twice what is was before. I.e. If intially the fundamental frequency was 50Hz, the new fundamental frequency would be 100Hz.

~H
 
  • #90
Alt+F4 said:
let me just quote this so it doesn't get lost

I answered this above in post #86 :smile:

~H
 
  • #91
Hootenanny said:
I answered this above in post #86 :smile:

~H
that is a different problem
 
  • #92
Alt+F4 said:
that is a different problem

Ahh, so it is. Perhaps it's time for me to take a break and get my fix of caffine :zzz:

~H
 
  • #93
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
That link isn't working.
 
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  • #94
sorry. http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99
 
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  • #95
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp99 Question 25

here was my plan divide the thing into 2 so 55 M on left, 55 M on right

do Tan 35 = X / 55

The Height = 38.5114
Huh? The motion is parabolic, not straight line.

Okay then i decided to just do

38.5114 = .5 * 9.8 * T^2
T = 2.80

Okay another way

(2)(9.8)* 38.5114 = V^2

V = 27.47405

so what i did was divide that be Cos 35 and got 33.53 why am i off?
(http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99; #25)

Treat x and y components separately and set up two equations for position as a function of time. Solve them together to find the initial speed.
 
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  • #96
Doc Al said:
Huh? The motion is parabolic, not straight line.


(http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp99; #25)

Treat x and y components separately and set up two equations for position as a function of time. Solve them together to find the initial speed.
so what are you pointing too,

I know X componet is

V Cos 35 t


Y componet

V sin T - .5*9.8 * t^2
 
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  • #97
Alt+F4 said:
so what are you pointing too,

I know X componet is

V Cos 35 t


Y componet

V sin T - .5*9.8 * t^2
That's right. You know the coordinates of the landing point at time t.
 
  • #98
so?

V Cos 35 t =110

110/ V cos 35 = T


and then substitute T in that V sin T equation?

and then the distance should equal 0 right/>
 
  • #99
Alt+F4 said:
so?

V Cos 35 t =110

110/ V cos 35 = T


and then substitute T in that V sin T equation?
That will work.

and then the distance should equal 0 right?
When the ball lands, its coordinates are x = 110 m; y = 0.
 
  • #100
A satellite is in circular orbit at a fixed radius from the center of the Earth and with a constant speed. Which one of the following statements is correct about the satellite?

(a) The acceleration is constant but the velocity is not.
(b) Both the acceleration and the velocity are constant.
(c) Neither the acceleration nor the velocity are constant.


Ans:C


WHy so? Constant Speed = Constant Velocity
 
  • #101
Okay so for a problem, say i wana find how far must u bring ur hands across the string to get a frequency of 350. How would u go about solving that
 
  • #102
Alt+F4 said:
A satellite is in circular orbit at a fixed radius from the center of the Earth and with a constant speed. Which one of the following statements is correct about the satellite?

(a) The acceleration is constant but the velocity is not.
(b) Both the acceleration and the velocity are constant.
(c) Neither the acceleration nor the velocity are constant.


Ans:C


WHy so? Constant Speed = Constant Velocity
Careful! Velocity and acceleration are both vectors--direction counts.
 
  • #103
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp99 Question 18

(5)(6) + (2) ( 20) - (x)(25) = 0

X = 2.8 why is it 2
 
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  • #104
well

Px = 0
Py = 0
 
  • #105
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp99 Question 18

(5)(6) + (2) ( 20) - (x)(25) = 0

X = 2.8 why is it 2
First things first. Momentum is a vector. Start by finding the total momentum of the 5-kg and 2-kg pieces.
 
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