well because it is a frictionless surface, m1 only has 1 relevant force acting on it, and that is of the rope. The force on it would be equal to the tension, so a = f/m.
Ugh, I am so confused by this. with the equation 2T - m2*g = m(-a), don't the m2*g and m(-a) cancel out, leaving just 2T = 0?
Im not sure what the equations are. I thought it'd be f=ma, so 2.95*9.8=28.91N, however this is not the answer. I am not sure what other equation to try
Sorry, this was a typo " x component = T2cos(55)" it should be
"x component = T2sin(55)"
Is that what you meant when you said I got the x and y components mixed up? I don't see anything wrong with the T1 components...sin(A)=opp/hyp, cos(A)=adj/hyp, that's right, right?
Anyway, I changed...
the part where I am having trouble is drawing the free-body for m2. The only force acting on it is (9.8*2.95) = 28.91N in the downward direction, right? But I'm not sure how the rope attached to the ceiling affects it's fall.
As for #2. I solved for the components of each. What I got was for...
Homework Statement
In the drawing, the rope and the pulleys are massless, and there is no friction. (m1 = 8.4 kg, and m2 = 2.95 kg.)
(a) Find the tension in the rope.
(b) Find the acceleration of the 8.4-kg block. (Hint: The larger mass moves twice as far as the smaller mass.)...