1. The problem statement, all variables and given/known data The system in the figure uses two masses m (I will call them m1 and m2) to raise a mass M. Consider that ropes are inextensible and that the mass of ropes and pulleys is zero. Find the acceleration of each body in terms of m,M,α and g. After that, indicate what is the minimum value of m so that M rise to a height yM in time t. I will call the x component of acceleration of mass m1, m2, and M: am1(x), am2(x) and aM(x) respectively. Pulley A is the one in direct contact with M. 3. The attempt at a solution My biggest issue was setting the reference system. I set x axes parallel to the floor and y axes positive downward. One constraint is that the length of the ropes are constant so that: length= ym1/sen(α) + 2yA + ym2/sen(α) and length of shorter rope= yM-yA , where ym1, ym2 and yA are the coordinates of masses m1, m2 and pulley A in the y system. This leads to: ym/sen(α)=-yA=-yM Y called the tension on the longer and shorter rope T1 and T2 respectively. Because pulley A is massless, 2T1=T2 , because anything else would mean an infinite acceleration of the pulley. Newton's equation, using my reference system (N is the normal force perpendicular to the plane and is equal to m*g*cos(α): m1*am1(x)=-N*sen(α) + T1*cos(α)=(-m*g*sen(α)+T1)*cos(α) m1*am2(y)=-N*cos(α)+m*g-T1*sen(α)=m*g*sen2(α) - T1*sen(α) m2*am2(x)=-m1*am1(x) m2*am2(y)=m1*am1(y) M*aM(y)=-T2=-2T1 I solved this equations for the accelerations: aM(y)= (2*g*sen(α)) / (M/m - 2) am(y)= (-2g*sen2(α)) / (M/m - 2) am(x)= (2*g*sen(α)*cos(α) / (M/m - 2) For the second part of the problem I integrated aM(y) twiced, and because it is constant, it leads to yM=aM(Y)*(t2/2) I solved this equation for m: m=(yM*M) / (t2*g*senα + 2yM) Something must be wrong with this last equation, because m converges to 0,5 for a constant time when yM tends to infinity, but I can't see where is the problem, I supposed that there issomething wrong with the reference system I used.