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Physics 99 Online
OpenStudy (x3_drummerchick):

Will give medals!! Can someone break down how to solve for m2? I'm so lost. My teacher wants us to break everything down into net y and net x component equations for each mass. ive enclosed a picture of the problem

OpenStudy (x3_drummerchick):

OpenStudy (x3_drummerchick):

OpenStudy (x3_drummerchick):

@mathstudent55 @mathmale

OpenStudy (irishboy123):

This is how to organise it |dw:1476388578653:dw| draw it, select a cordinate system, and then write some equations

OpenStudy (x3_drummerchick):

@Irishboy123 is that for solving m2?

OpenStudy (x3_drummerchick):

@Nnesha @jabez177

OpenStudy (x3_drummerchick):

my teacher wants the m1 x and y components, m2 components and m3 components

OpenStudy (x3_drummerchick):

@SapphireMoon

OpenStudy (sapphiremoon):

Sorry, @x3_drummerchick haven't done that unit yet.. (Should have done it two weeks ago but eh)

OpenStudy (x3_drummerchick):

poop :( thank you anyway

OpenStudy (irishboy123):

you can solve for \(m_2 \) from the first equation only, which comes from resolving the forces in the horizontal direction ie \(- T_1 \cos \alpha + T_2 \cos \beta = 0\) IOW \(- m_1 g \cos \alpha + m_2 g \cos \beta = 0\)

OpenStudy (danjs):

You have both angles, alpha and beta, and masses m1 and m3. The X direction force sum shows you , it is in equilibrium and not accelerating so the net force is zero. \[\large \sum F _{y}=-m _{1}g*\cos(\alpha)+m _{2}g*\cos(\beta)=0\] \[m _{1}g*\cos(\alpha) = m _{2}g *\cos(\beta)\] \[\large m _{2}=\frac{ m _{1}*\cos(\alpha) }{ \cos(\beta) }\]

OpenStudy (janet8485858):

can anyone help me please.

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