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
  • #36
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 17 , um you so 45/270 = Omega = .16667

How do i go about knowing the lap time? DO i find the Area? and would that be The change in X
 
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  • #37
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05

Question 14 , i know i am doing it right but i can't get the answer

so all it is is

Force of Friction - F Cos 64 = 0
This is correct so far.

(.1)(9.80(7) - F (cos 64)
This is not. Hint: Did you answer Question 13?
 
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  • #38
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 17 , um you so 45/270 = Omega = .16667

How do i go about knowing the lap time? DO i find the Area? and would that be The change in X
Snap out of it! :smile: Try using Distance = speed * time.
 
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  • #39
Doc Al said:
Snap out of it! :smile: Try using Distance = speed * time.
what is Distance? O man i don't even know anymore
 
  • #40
last 2 of the day, then I am hitting the bed since I am losing it

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 22 and 23


So i know that at Point H it takes .75 Sec to get there but how do i figure the time it takes to land back at 4m. and i guess if i get that then 23 will help itself
 
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  • #41
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

QUestion 17,

So

-Fpush Cos 30 - Force of Friction = 0

Fpull Cos 30 - Force Of Friction = 0

Fpull should be larger but it is not, is my math wrong?
 
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  • #42
The sandbag has vertical speed zero at the top (H) .

how far above the thrower is this?

method a) v_o t + 1/2 a t^2 ?

method b) average vertical speed , for how long upward?
 
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  • #43
F_push is not equal to F_pull .

You'll need to look at the CAUSE of the friction Force to determine that
(not just the observable effect of the TOTAL Force).
 
  • #44
Doc Al said:
This is correct so far.


This is not. Hint: Did you answer Question 13?
yes i answerd question 13 since it is F sin Theta - mg + normal = 0

Normal = mg - F sin theta which becomes Normal is smaller than mg

Edit: i answerd my own question
 
  • #45
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 Question 17 , um you so 45/270 = Omega = .16667

How do i go about knowing the lap time? DO i find the Area? and would that be The change in X
So what would be the circular Distance for this bad boy? It is a circle so 360 degrees am i thinking too much?
 
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  • #46
degrees are usually used to measure ANGLES, not DISTANCES.
hint : see what units the velocity (or acceleration) is measured in

Are you asking how far it is around a circle?
The distance around a hexagon (equilateral triangles of length R) is 6R ...
 
  • #47
Alt+F4 said:
So what would be the circular Distance for this bad boy? It is a circle so 360 degrees am i thinking too much?
okay well this is what i did, i don't even know what it means but

A= pi R^2 = 229022

229022/ (45 * 270/2))
 
  • #48
lightgrav said:
degrees are usually used to measure ANGLES, not DISTANCES.
hint : see what units the velocity (or acceleration) is measured in

Are you asking how far it is around a circle?
The distance around a hexagon (equilateral triangles of length R) is 6R ...
well omega is 45/270 = .1666

I need to find the time, the answer is 37.7
 
  • #49
i got it, i forgot there was such a thing as circumference
 
  • #50
Did you get the box push vs pull when you realized the answer to prob.13?
 
  • #51
lightgrav said:
Did you get the box push vs pull when you realized the answer to prob.13?
well i understood it from the beginning, i just wanted to know if there was an equation in case it is asked on the final to find one of them
 
  • #52
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 I still don't get how to do 21 and 23, so i got .7536 seconds to reach at its highest

Actually i found the answer to 22 which was 4.3 4.0 - 4.3 = .3 meters left to fall

. 3 = .5 9.8 T^2

T = .2474 + .7536 = 1.001

Now 23
 
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  • #53
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 I still don't get how to do 21 and 23, so i got .7536 seconds to reach at its highest

Actually i found the answer to 22 which was 4.3 4.0 - 4.3 = .3 meters left to fall

. 3 = .5 9.8 T^2

T = .2474 + .7536 = 1.001

Now 23
For 23, you have two choices:

you can do it with the equations of kinematics ([itex] v_x[/itex] stays constant. For [itex] v_y[/itex] you may use [itex] v_y(t) = v_{y,initial} - g t [/itex]. Then the final speed is given by Pythagora's theorem.

OR you could use conservation of energy
 
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  • #54
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05 I still don't get how to do 21 and 23, so i got .7536 seconds to reach at its highest

Actually i found the answer to 22 which was 4.3 4.0 - 4.3 = .3 meters left to fall

. 3 = .5 9.8 T^2

T = .2474 + .7536 = 1.001

Now 23
What's the problem with 23? To find speed at some point, use kinematics or energy conservation. (Horizontal component of velocity remains constant.)
 
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  • #55
nrqed said:
For 23, you have two choices:

you can do it with the equations of kinematics ([itex] v_x[/itex] stays constant. For [itex] v_y[/itex] you may use [itex] v_y(t) = v_{y,initial} - g t [/itex]. Then the final speed is given by Pythagora's theorem.

OR you could use conservation of energy
so V y intitial = 7.5 Cos 80 = 1.30236
so v (t) = 1.30236 - (9.8)(.2474)
V X intial = 7.5 sin 80 = 7.386

Then what? i am not getting how to use the forumla
 
  • #56
Doc Al said:
What's the problem with 23? To find speed at some point, use kinematics or energy conservation. (Horizontal component of velocity remains constant.)
so horizontal componet is 7.5 cos 80 = 1.3
 
  • #57
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp05

Question 25

So what i did was 4 cos 22 = 3.7087 M/s

Time * Velocity = Distance

(120)(3.7087) = 445.04

Ans: 450, did they just round?
 
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  • #58
Alt+F4 said:
so V y intitial = 7.5 Cos 80 = 1.30236
so v (t) = 1.30236 - (9.8)(.2474)
V X intial = 7.5 sin 80 = 7.386

Then what? i am not getting how to use the forumla
You must use sine for the y component and cos for the x component (because of the way the chose their angle).

In the end you are getting the x and y components of the final velocity vector. To find the final speed, just calculate the magnitude of the final velocity vector (so use [itex] v_f = {\sqrt{v_{x,f}^2+v_{y,f}^2}}[/itex])
 
  • #59
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04

Question 9

Well i found the weight of the other mass which is also 2.5, so all i did was add 15 N + 15N = 30 which is the answer is this the correct way or did i luck out
 
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  • #60
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp04 Question 11, why are they equal there is a verticle component for the second box which is mgsin 45 while for the first it is 0
 
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  • #61
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 verticle 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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  • #62
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 postive some times it is negative
 
  • #63
Hootenanny said:
Well, how do you find the vertical component?

~H
well verticle component is just if there is an angle then you have one
 
  • #64
Alt+F4 said:
well verticle 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
 
  • #65
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
 
  • #66
Alt+F4 said:
Normal Force is acting upward

Yep, but what would happen if there was a net force acting upwards on a block?

~H
 
  • #67
Hootenanny said:
Yep, but what would happen if there was a net force acting upwards on a block?

~H
then it would equal
 
  • #68
Alt+F4 said:
then it would equal

What? Using Newton's second law, what would happen if there was a net upwards force?

~H
 
  • #69
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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  • #70
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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