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Ac3
 one year ago
In the figure below, a tin of antioxidants (m1 = 1.3 kg) on a frictionless inclined surface is connected to a tin of corned beef (m2 = 2.3 kg). The pulley is massless and frictionless. An upward force of magnitude F = 5.9 N acts on the corned beef tin, which has a downward acceleration of 5.4 m/s2. What are the tension in the connecting cord and angle β?
Ac3
 one year ago
In the figure below, a tin of antioxidants (m1 = 1.3 kg) on a frictionless inclined surface is connected to a tin of corned beef (m2 = 2.3 kg). The pulley is massless and frictionless. An upward force of magnitude F = 5.9 N acts on the corned beef tin, which has a downward acceleration of 5.4 m/s2. What are the tension in the connecting cord and angle β?

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IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0of course have you drawn any free body diagrams? do you have your own thoughts on this?

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0Finding the tension like this gives 4.22. However that's wrong and the reason I know it's wrong is because another one with very similar numbers has the Tension as 2.6

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0Now how do I find the angle?

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443919249971:dw

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0n is the normal force from the surface on the can

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Explain to me how you find tension earlier?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Because you need that for angle.

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0T = 4.243N is good you need \(\beta\), right?

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0tension is found through that equation

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0obviously so you have for the upper block \(T + mg \sin \beta = ma \) ?? do you see that

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443920229507:dw

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0dw:1443920344245:dw

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0OHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0wow i can't believe I couldn't figure that out

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0I even drew that exact picture out as well

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0thanks Irishboy saving me again

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You mean for the upper block? \[mgsin \betaT=ma\]

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0@Ac3 blows me away too, every time. good luck!

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0no it is indeed +T not T, that's because they're both going down @Shalante

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0If its going down, then it would start with mg is first. But you got the answer right?

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0@IrishBoy123 I use M1 for the mass correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0You need to derive a lot.

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0I got the angle wrong. Must have messed up the calculation. I got 28 for the positive angle and 61.9 for the negative angle

IrishBoy123
 one year ago
Best ResponseYou've already chosen the best response.0@Ac3 run with @Shalante see where you get to. tag me tomorrow otherwise, gtg. good luck!

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0what did you get for the angle @Shalante

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Tell me how you got tension first

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0plug in all values and solve for T. That's how you get tension.

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0I guess plugging in all the values at once in the calculator gives you the right answer. lol

Ac3
 one year ago
Best ResponseYou've already chosen the best response.0Thanks a lot @IrishBoy123.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Let me guess the angle.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Nvm. Someone tagged me.
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