I'm guessign at that point I could then use:
Tnet = [ Tension2 - Tension1]*radius = .5 m r^2 ∝ ?
Tension2*cos(22) - Magsin(30) - ( - Tension sin(22) + mgcos(30) )μ = 0
Put this equation in terms of Tension2
solve the hanging block equation for t1
Mb a = T1 - Mb g
Tension1= Mb a + Mb g
The y component would be sin() of it... the only thing i am confused about now
Now I have ay and ax though? Look I really appreciate your help but going through this step by step has become pretty tedious, what do you think the equation should look like?
System comprised blocks, a light frictionless pulley and connecting ropes (see diagram). The 9.0kg block is on a perfectly smooth horizontal table. The surfaces of the 12kg block are rough, with μk = .2 between the two blocks. If the 5.0 kg block accelerates downward when it...
This would be my equation for may,
may * acceleration = Normal Force + Tension sin(22) - mgcos(30)
I guess I could only solve for the normal force within the whole system of equations, or I could put this in terms of the normal force and solve for this as Normal force in the other equation? If...
Okay, so here is what I think the force diagram is for the block, ignore the other stuff though.
So my equation for part 1 is Tension*cos(22) - Ma g sin(30) - ( Ma g cos(30) - Tension*sin(22) )μ = 0
sometimes I think about it such as the equation is parallel and perpendicular.
Would this be...
Oh, okay.. i see what your saying.. wouldnt it be something like:
100gcos(30) - (Tension2*sin(22) + 100gcos (30)
The tension and normal force acting in one direction, with the force of direction in the other? other wise I am a bit lost here haha
Sorry, Not sure what I ment by that. I put 100gcos(30) - Tension2*sin(22) - Ma A
in because I figured that when I calculated the normal force I would need to account of Acceleration as well since it is now moving.. now that I say that it doesn't make a ton of sense to me.. should I have just...
Woops, Thanks for the fast reply.
Yeah, that's what I was implying. Looking at the diagram, I was trying to account for the Y side.. and figured that the Y would be changing as well as it is not at rest..
I edited the original post, I stupidly forgot the G.
Tension2*cos(22) - Magsin(30) - (...
100 kg box is held in place on a ramp that rises at 30° above the horizontal. There is a massless rope joint to the box that makes a 22° angle above the surface of the ramp. Coefficients of friction between the box and the surface of the ramp are μk = 0.40 and μs = .60. The...