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catsrule332
Group Title
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.
 2 years ago
 2 years ago
catsrule332 Group Title
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.
 2 years ago
 2 years ago

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imalittletoyrobotfromthefuture Group TitleBest ResponseYou've already chosen the best response.0
hi abby
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
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?
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
The object is swinging in a vertical circle.
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
Ah. Perfect.
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
Now, we need to know where the ball is located at the instance shown. Can you give me an angle, say 30 degrees counterclockwise from the horizontal?
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
dw:1332714065917:dwSomething like this would help.
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
Here I will give you the a picture:dw:1332714090745:dw
 2 years ago

srinidhijha Group TitleBest ResponseYou've already chosen the best response.1
u should hv stated earlier that ball is swinging in circle
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
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\[F_c = m{v^2 \over r} = m r \omega^2 = T\]
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
Understand?
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
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 xdirection and g acts in the ydirection.
 2 years ago

srinidhijha Group TitleBest ResponseYou've already chosen the best response.1
net forc is 4.03 N by Pythagoras theorem root ( 4^2 + .0.5^2) = 4.03 N
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
Answers aren't helpful. Please try to explain your methodology. Refer to the new CoC.
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
http://openstudy.com/codeofconduct
 2 years ago

srinidhijha Group TitleBest ResponseYou've already chosen the best response.1
please correct me I want an answer too Its a humble request please
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
srinidhijas method using the Tmg=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.
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
The Force vector and the acceleration vector will have the same direction. Therefore, \[F = a_n \cdot m = \sqrt{T^2 + F_g^2}\]
 2 years ago

srinidhijha Group TitleBest ResponseYou've already chosen the best response.1
dw:1332714774000:dw
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
T  mg = ma is not valid here.
 2 years ago

srinidhijha Group TitleBest ResponseYou've already chosen the best response.1
yes u did not said earlier tht it was swingin in a circle atsrule
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
alright, sorry for not making that clear. eashmore the last equation you posted makes sense.
 2 years ago

eashmore Group TitleBest ResponseYou've already chosen the best response.0
In the xdirection, we have a tension force and only a tension force. T = 4N In the ydirection, 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 ydirection\[F_g = m \cdot a_y = m \cdot g\] We can use these two equations to find the acceleration in the x and ydirections. Then, using the same methodology as we did to find the net force, we can find the net acceleration as\[a = \sqrt{~ a_x^2 + g^2}\]
 2 years ago

catsrule332 Group TitleBest ResponseYou've already chosen the best response.0
Thank you both! I understand it now.
 2 years ago
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