anonymous
  • anonymous
The mass of the ball is 0.050kg, and at the instant shown the value of the tension is 4N. Find the net force on the ball and find the acceleration vector (give the x and y components). *T is the tension in the string and Fg is the gravitational force.
Physics
schrodinger
  • schrodinger
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anonymous
  • anonymous
hi abby
anonymous
  • anonymous
We need to better understand the orientation of the system. Is the object being pull across the ground, it the object hanging, is the object rotation? If it is rotating, what plane is the rotation in? Horizontal or vertical?
anonymous
  • anonymous
The object is swinging in a vertical circle.

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anonymous
  • anonymous
Ah. Perfect.
anonymous
  • anonymous
Now, we need to know where the ball is located at the instance shown. Can you give me an angle, say 30 degrees counter-clockwise from the horizontal?
anonymous
  • anonymous
|dw:1332714065917:dw|Something like this would help.
anonymous
  • anonymous
Here I will give you the a picture:|dw:1332714090745:dw|
anonymous
  • anonymous
u should hv stated earlier that ball is swinging in circle
anonymous
  • anonymous
Now we can work it. Notice that tension and gravity act orthogonally to each other. Therefore, the tension force is equal to the centripetal force\[F_c = m{v^2 \over r} = m r \omega^2 = T\]
anonymous
  • anonymous
Understand?
anonymous
  • anonymous
The net acceleration can be expressed as\[a_n = \sqrt{~a_c^2 + g^2 } = \sqrt{~ \left(v^2 \over r \right)^2 + g^2 }\] Notice that \(a_c\) acts in the x-direction and g acts in the y-direction.
anonymous
  • anonymous
net forc is 4.03 N by Pythagoras theorem root ( 4^2 + .0.5^2) = 4.03 N
anonymous
  • anonymous
Answers aren't helpful. Please try to explain your methodology. Refer to the new CoC.
anonymous
  • anonymous
anonymous
  • anonymous
please correct me I want an answer too Its a humble request please
anonymous
  • anonymous
srinidhijas method using the T-mg=ma seems to make sense, and seems like it would be correct. Eashmore I don't understand your method completely, if you could try to explain it a bit more that might help.
anonymous
  • anonymous
The Force vector and the acceleration vector will have the same direction. Therefore, \[F = a_n \cdot m = \sqrt{T^2 + F_g^2}\]
anonymous
  • anonymous
|dw:1332714774000:dw|
anonymous
  • anonymous
T - mg = ma is not valid here.
anonymous
  • anonymous
yes u did not said earlier tht it was swingin in a circle atsrule
anonymous
  • anonymous
alright, sorry for not making that clear. eashmore the last equation you posted makes sense.
anonymous
  • anonymous
In the x-direction, we have a tension force and only a tension force. T = 4N In the y-direction, we have a gravitational force and only a gravitational force. Fg = 0.005 * 9.81. The magnitude of these two forces is found from Pythagorus theorem\[F = \sqrt{~ T^2 + F_g^2} = \sqrt{~ 4^2 + (0.05 \cdot 9.81)^2}\] To find the acceleration in each direction, realize that the tension force is caused by the centripetal force. \[T = m \cdot a_x\]and that the gravitational force causes an acceleration in the y-direction\[F_g = m \cdot a_y = m \cdot g\] We can use these two equations to find the acceleration in the x and y-directions. Then, using the same methodology as we did to find the net force, we can find the net acceleration as\[a = \sqrt{~ a_x^2 + g^2}\]
anonymous
  • anonymous
Thank you both! I understand it now.

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