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
  • #1
Alt+F4
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Okay, I am back for one more time. Last thread of the semster. Thanks for everyone's help for the other 3.


Question 6

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06


How would you go about finding this problem

I know torque = FrSinTheta
 
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  • #2
Draw the diagram showing the force on the valve stem due to gravity. What is the angle between the force on the valve stem and the radius from the axle to the valve stem? Do you see why the answer to question 5 also involves this angle?
 
  • #3
If you know the equation for torque (a vector quantity), then all you have to do is find the magnitude.
 
  • #4
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 2, I don't get why it is true

V ave for A = -20 m/s
V ave For B = 10 m/s
V ave For C = 40 m/s

(10+-20+40)/(3) does not equal 0
 
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  • #5
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 8

Change in Y is 1

So what i was thinking of doing is

Actually i have no idea

I think the formula i should be using is V^2 = V0^2 + 2a(change in X)

A = 9.8
V0 = 12sin65
so i found V to be 11.74

Then i just used

V = V0 + at
which gave me .084 sec

Any help, thanks
 
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  • #6
O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 21
 
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  • #7
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 2, I don't get why it is true

V ave for A = -20 m/s
V ave For B = 10 m/s
V ave For C = 40 m/s

(10+-20+40)/(3) does not equal 0
Your logic is faulty. Think about this: during period A you earned $100/hour; during period B you earned $1/hour. Are you sure that your average earning rate is (101)/2 = $50.50/hour? What if period A lasted 1 hour but period B lasted 10 years?

The right way to find average velocity is displacement/time.
 
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  • #8
Doc Al said:
Your logic is faulty. Think about this: during period A you earned $100/hour; during period B you earned $1/hour. Are you sure that your average earning rate is (101)/2 = $50.50/hour? What if period A lasted 1 hour but period B lasted 10 years?

The right way to find average velocity is displacement/time.
I want to thank you for all your help, I aced all the exams, just this final and i am done. Thanks
 
  • #9
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 8

Change in Y is 1

So what i was thinking of doing is

Actually i have no idea

I think the formula i should be using is V^2 = V0^2 + 2a(change in X)

A = 9.8
V0 = 12sin65
so i found V to be 11.74

Then i just used

V = V0 + at
which gave me .084 sec

Any help, thanks
Hint: What's the horizontal velocity and acceleration?
 
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  • #10
Doc Al said:
Hint: What's the horizontal velocity and acceleration?
Horizontal Velocity = 12 Cos 65 = 5.0714

Acc = -9.8

What is confusin me is the Distance factor The Hoop is 3 meters what do i do with that
 
  • #11
Alt+F4 said:
O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 21
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)?
 
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  • #12
Alt+F4 said:
Horizontal Velocity = 12 Cos 65 = 5.0714
Right (in m/s).

Acc = -9.8
Wrong! (What's the horizontal acceleration?)

What is confusin me is the Distance factor The Hoop is 3 meters what do i do with that
What you do with it is ignore it. You don't need it. Focus on the horizontal component of the motion. You have the speed and the distance.
 
  • #13
Doc Al said:
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)?
am i making it harder by assuming Vectors and Just figuring it out there cause the triangle is 45 45 90, i can find the angle then from there i can find the speed
 
  • #14
Alt+F4 said:
am i making it harder by assuming Vectors and Just figuring it out there cause the triangle is 45 45 90, i can find the angle then from there i can find the speed
You can certainly use vectors, but that's more work than needed. I don't see a 45-45-90 triangle.
 
  • #15
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 9

So i got 1.97 Sec for 8

so ithinkin of using X = X0 + V0T + .5 a T^2

But that doesn't work
 
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  • #16
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 9

So i got 1.97 Sec for 8
OK.

so ithinkin of using X = X0 + V0T + .5 a T^2

But that doesn't work
Why not? It works for me. (Be sure you are using vertical components and proper signs.)
 
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  • #17
Alt+F4 said:
O yes, These types of problems, Can somone just explain how to exactly do it, there are like 20 more on the other exams. I never really got how. Thanks

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/sp06

Question 21


Okay so let's see if i got what you are saying

Vave = X/T

1.8 = 55/ T

T = 30.555 sec

V ave = 75 / 30.5555

V ave = 2.5

Is this the correct way?
 
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  • #18
Alt+F4 said:
Okay so let's see if i got what you are saying

Vave = X/T

1.8 = 55/ T

T = 30.555 sec

V ave = 75 / 30.5555

V ave = 2.5

Is this the correct way?
Looks good to me. The person's speed with respect to the water is 2.5 m/s. (If you wanted to find the person's speed with respect to the shore, you'd have to add the cross-stream (2.5 m/s) and down-stream (1.8 m/s) components--using vector addition--to find his total speed. But you weren't asked that.)
 
  • #19
okay question 5 and 6

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06


Was question 5 basically you know it or you don't type of thing?

And free body digram would be

Stem Valve Pointing down = mg = .245

T = F r sin theta
 
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  • #20
Doc Al said:
Looks good to me. The person's speed with respect to the water is 2.5 m/s. (If you wanted to find the person's speed with respect to the shore, you'd have to add the cross-stream (2.5 m/s) and down-stream (1.8 m/s) components--using vector addition--to find his total speed. But you weren't asked that.)


So like (1.8^2) + (2.5^2) = X^2
 
  • #21
Alt+F4 said:
okay question 5 and 6

http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06


Was question 5 basically you know it or you don't type of thing?

And free body digram would be

Stem Valve Pointing down = mg = .245

T = F r sin theta
Careful with your [itex]\sin\theta[/itex]. In the equation for torque, [itex]\theta[/itex] is the angle between the radius vector (r) and the force vector (in this case, mg). That's not the same as the angle [itex]\theta[/itex] shown in the diagram (but they are related, of course).

So what's your answer?
 
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  • #22
Alt+F4 said:
So like (1.8^2) + (2.5^2) = X^2
Exactly...
 
  • #23
Doc Al said:
Careful with your [itex]\sin\theta[/itex]. In the equation for torque, [itex]\theta[/itex] is the angle between the radius vector (r) and the force vector (in this case, mg). That's not the same as the angle [itex]\theta[/itex] shown in the diagram (but they are related, of course).

So what's your answer?


Okay so For 5 it is 90 degree so 90 - 24 = 66 and (.025)(9.8)(.3) *sin 66

So where exactly is 66 degrees? is it the one that is drawn? Thanks
 
  • #24
Alt+F4 said:
Okay so For 5 it is 90 degree
That's the answer to #5: the torque will be minimum (equal to zero!) when theta = 90 degrees.

so 90 - 24 = 66 and (.025)(9.8)(.3) *sin 66
That's the answer for #6. When the angle shown is 24; the angle between force and radius is 90 - 24 = 66 degrees.

So where exactly is 66 degrees? is it the one that is drawn?
66 degrees will be the angle between the force and the radius; or, in other words, between the vertical axis and the radius vector (since the force, gravity, acts vertically). (The diagram shows the angle between the horizontal axis and the radius vector.)
 
  • #25
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06

Question 20

Impulese = F ave Change in time = Change in Momentum

Fave ( .07) = ( .125)(Vf ) = (.125)(7)


How do i find Vf

I know it bounces back 2 m

so what i did was

(2) ( 9.8) = .5 (V^2)

V = 6.26 but that doesn't work


Edit: i guess i didnt read the part about it going downward so

( M )( Vf) + (M) (Vi)

Which does work
 
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  • #26
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/sp06

Question 20

Impulese = F ave Change in time = Change in Momentum

Fave ( .07) = ( .125)(Vf ) = (.125)(7)


How do i find Vf

I know it bounces back 2 m

so what i did was

(2) ( 9.8) = .5 (V^2)

V = 6.26 but that doesn't work


Edit: i guess i didnt read the part about it going downward so

( M )( Vf) + (M) (Vi)

Which does work
I assume this is question 13, not 20.

The initial velocity (just before contact) is given as -7 m/s; you calculated the final velocity (just after contact) as +6.26 m/s. So the change in momentum is m(Vf - Vi) = m(6.26 - -7) = m(13.26). Now you can divide by the time to find the average force.
 
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  • #27
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa05

Question 9

I have no idea i guess you would end up using this formula

Fg = Gm1M2 / R^2 but what would be the second mass?
 
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  • #28
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa05

Question 21

So i drew out free body diarm, i got the tension for previous which was 24N


-T - mgsin30 + UkMg = ma

-24 - (X)(9.8)sin 30 + (.2)(X)(9.8) = (X)(-5)

I end up getting 11.65
 
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  • #29
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05

Question 26

So what i did was

(1.5 * 10^5)(-4) = (1.5*10^6)(X) but that don't make sense

I know since it bounces off it will be negative
 
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  • #30
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa05

Question 9

I have no idea i guess you would end up using this formula

Fg = Gm1M2 / R^2 but what would be the second mass?
Hint: Consider the acceleration and apply this formula (for gravity) along with Newton's 2nd law.
 
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  • #31
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam1/fa05

Question 21

So i drew out free body diarm, i got the tension for previous which was 24N
OK.

-T - mgsin30 + UkMg = ma

-24 - (X)(9.8)sin 30 + (.2)(X)(9.8) = (X)(-5)

I end up getting 11.65
Careful with your signs: The tension and acceleration point up the ramp; friction and weight point down.
 
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  • #32
Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys101/spring06/prep2a.pl?practice/exam2/fa05

Question 26

So what i did was

(1.5 * 10^5)(-4) = (1.5*10^6)(X) but that don't make sense

I know since it bounces off it will be negative
Think about it qualitatively, don't just reach for a formula. You know that the total momentum (which is a vector) will be the same for any kind of collision. But which collision gives more momentum to the iceberg? Hint: How does the momentum of the ship (post collision) differ in each case?
 
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  • #33
Doc Al said:
Think about it qualitatively, don't just reach for a formula. You know that the total momentum (which is a vector) will be the same for any kind of collision. But which collision gives more momentum to the iceberg? Hint: How does the momentum of the ship (post collision) differ in each case?
ya i understood it was gona be less than but i wana know in case he asks to figure out its Velocity
 
  • #34
For question 26:

To solve for the speed of the iceberg, set the initial momentum of the system (originally, just the ship is moving) equal to the final momentum of the system (both ship and iceberg). The only unknown will be the speed of the iceberg.
 
  • #35
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

(.1)(9.80(7) - F (cos 64)

F = 15.648
 
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