Two problems i just can't get

  • Thread starter prettyinpink
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In summary, the first problem involves calculating the mass of a person on a playground carousel based on changes in angular velocity. The second problem involves finding the tension in a support cable and the force exerted on the lower end of a boom by a hinge. Equations such as \sum F_x = 0 and \sum F_y = 0 can be used to solve for unknown forces and moments. It is important to break forces into horizontal and vertical components and use angles to determine the direction of the forces.
  • #1
prettyinpink
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A playground carousel is free to rotate about its center on frictionless bearings, and air resistance is negligible. The carousel itself (without riders) has a moment of inertia of 120 kgm2. When one person is standing at a distance of 1.40 m from the center, the carousel has an angular velocity of 0.500 rad/s. However, as this person moves inward to a point located 0.740 m from the center, the angular velocity increases to 0.640 rad/s. What is the person's mass?


A wrecking ball (weight = 4500 N) is supported by a boom, which may be assumed to be uniform and has a weight of 3100 N. As the drawing shows, a support cable runs from the top of the boom to the tractor. The angle between the support cable and the horizontal is 32°, and the angle between the boom and the horizontal is 48°.
(a) Find the tension in the support cable.

(b) Find the magnitude of the force exerted on the lower end of the boom by the hinge at point P.


For the first problem I tryed doing IW=IW and didn't get the right answer.
For the seconde problem i understand the froces i just don't know how to put them into the equation. (https://www.physicsforums.com/showthread.php?p=811650)
 
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  • #2
You should have used the homework thread and specified what you have done. Try to show what you did, explain the equations you are using, etc.
 
  • #3
Sorry about that I thought I did use the homework thread. For the first problem I did WI=I1W1+I2W2 and so with the number is would look like this (120)(.5)=(120+m0.740 squared)(.640) ...i know I is mrsquared.
For the seconde I drew all the forces as included in the link above when someone else had this problem but didn't know how to set up the equation to slove for the answers.
 
  • #4
prettyinpink said:
Sorry about that I thought I did use the homework thread. For the first problem I did WI=I1W1+I2W2 and so with the number is would look like this (120)(.5)=(120+m0.740 squared)(.640) ...i know I is mrsquared.

You applied something on the right hand side of the red equation that you didn't apply on the left hand side. Do so, and you should obtain the correct answer.
 
  • #5
thxs

thxs ill check that out any idea on the seconde problem
 
  • #6
prettyinpink said:
thxs ill check that out any idea on the seconde problem

The idea is to visit the link you gave to the other thread with the same problem and investigate it once again. If you get stuck, be specific about what you got stuck with. :smile:
 
  • #7
I don't understand which forces to use to find the tension and how to use and equation to find the tension
 
  • #8
Do a balance of forces, both horizontal and vertical. Then do a balance of moments (or torques). (from post) ? confused
 
  • #9
Static equilibrium of rigid bodies

Hi prettyinpink,

For this question, u may need to use the equations
[tex]\sum F_x = 0 , \sum F_y = 0 and \sum \tau = 0 [/tex].
Let the tension in the support cable be T and the normal reaction force at P be R.
Then, using [tex]\sum F_y = 0 [/tex], u'll get T sin 32 + R sin 48 - 4500 - 3100 = 0.
Using [tex]\sum F_x = 0 [/tex], u'll get T cos 32 + R cos 48 = 0.
Solve the 2 equations simultaneously and u can get the values of T and R. T is needed for part (a) while R is needed for part (b).

Hope that helps.
 

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  • #10
i certainly will try the equations out today thxs much!
 
  • #11
NTUENG said:
Hi prettyinpink,

For this question, u may need to use the equations
[tex]\sum F_x = 0 , \sum F_y = 0 and \sum \tau = 0 [/tex].
Let the tension in the support cable be T and the normal reaction force at P be R.
Then, using [tex]\sum F_y = 0 [/tex], u'll get T sin 32 + R sin 48 - 4500 - 3100 = 0.
Using [tex]\sum F_x = 0 [/tex], u'll get T cos 32 + R cos 48 = 0.
Solve the 2 equations simultaneously and u can get the values of T and R. T is needed for part (a) while R is needed for part (b).

Hope that helps.

NTUENG, We appreciate your helping out with homework questions, but please avoid doing a lot of the OP's work for them. In your post above, it would have been better for you to ask more tutorial questions, rather than doing the math for the component forces for the student. Your intentions are good, but you need to understand that doing the work for the student does not help them learn as much as guiding them towards discovering out how to do the work.
 
  • #12
no go

the equations weren't right anyway and i still have no idea how to write the equations...
 
  • #13
Yes I still need help with the equations I don't understand how to write the forces in an equation. The equations that were given from above produced a negative answer and I did check my work.
 
  • #14
Okay, for the second question you need to split the forces acting into horizontal and vertical components. Can you do that?
 
  • #15
I know there has to be an x and y direction but i don't know what angles to use and wether you add or subtract the numbers.
 
  • #16
never mind...i sloved it
 

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