Where Did I Go Wrong in Solving This Physics Problem?

[BIGEYE]

Appreciate if someone could take the time to look at the following problem, my answers, and can point out where I have went wrong. Also, part 4, I think more info. is needed to determine this, but I may be wrong. Anyway, here goes:

A hoist motor is attached to a support beam as shown in the diagram. The hoist lifts a load of 3.5kg, the load starts from rest and has a final velocity of 2 m/s after 4 seconds. The motor can develop a torque of 4 Nm.
Determine:
i) The linear acceleration of the mass
ii) The angular acceleration of the motor
iii) The force on the beam at M when the hois is lifting the load
iv) The reaction forces RA and RB on the beam supports A and B

My attempt at solutions:
i) Linear acceleration (a) = (final velocity - initial velocity) / time
= (2 - 0)/4
= 0.5 m/s/s

ii) Work done in 4 seconds = 0.5mv^2
= 0.5 * 3.5 * 2^2
= 7 Joules
Angle of rotation = Work / Torque
= 7/4
= 1.75 radians in 4 seconds
Angular acceleration = 0.4375 radians/s/s

iii) Force = mass * acceleration = 3.5 * (9.81 + 0.5)
= 36.085 N

iv) Don't know how to tackle this part as it is.

Sketch can be found here
http://img443.imageshack.us/img443/5980/q2mplg7.jpg

 

[member 553083]

For part i) you have correctly calculated the linear acceleration.For part ii) you have correctly calculated the work done, however you have incorrectly calculated the angular acceleration. The angular acceleration is equal to the change in angular velocity divided by the time interval. Since the angular velocity starts at 0 rad/s and ends at 1.75 rad/s after 4 seconds, the angular acceleration will be equal to (1.75 - 0)/4 = 0.4375 rad/s^2. For part iii) you have correctly calculated the force on the beam at M.For part iv) the reaction forces RA and RB on the beam supports A and B can be calculated using the equation of equilibrium:F_x=RA-RB=0F_y=RA+RB-3.5*(9.81 + 0.5)=0Solving these two equations yields RA = 17.54 N and RB = 17.54 N.