Final Exam Q6: Finding Problem with Torque Equation

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Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04 Question 11, why are they equal there is a vertical component for the second box which is mgsin 45 while for the first it is 0

Well, how do you find the vertical component?

~H
 
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Stupid question, but i get soo confused when do i know the say acceleration will be negative in this equation like Friction - Tension = ma

Is there key words in the problem or what cause sometimes it is positive some times it is negative
 
Hootenanny said:
Well, how do you find the vertical component?

~H
well vertical component is just if there is an angle then you have one
 
Alt+F4 said:
well vertical component is just if there is an angle then you have one

What would happen if there was an unbalanced force acting upwards?

HINT: What other force is acting upwards in BOTH cases.

~H
 
Hootenanny said:
What would happen if there was an unbalanced force acting upwards?

HINT: What other force is acting upwards in BOTH cases.

~H
Normal Force is acting upward
 
Hootenanny said:
Yep, but what would happen if there was a net force acting upwards on a block?

~H
then it would equal
 
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa03

Question 8, why couldn't it not have been done like this 1.4/9.8 = .14
 
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Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa03

Question 8, why couldn't it not have been done like this 1.4/9.8 = .14

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
 
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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
 
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
 
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
 
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp03

Question 23

I did centripetal Acceleration * weight

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

Question 23

I did centripetal 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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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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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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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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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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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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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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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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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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WHen it says you Double The Frequency does it Actually mean Multiply it by 2 or does it mean Half it?
 
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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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
 
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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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