Recent content by DukeJP2010

Ok, since the direction of travel of the mass is always perpendicular to Ft, then Ft can't do any work. Does the rest of it look OK?

Homework Statement A simple pendulum (sting has no mass, no air resistance) has a weight on the end of mass m (unknown). The string has a length A horizontal (and always horizontal) force, F, acts on the weight, but it moves so slowly that acceleration is 0. a)What is the work done by force...

Once again, that you for your help. Topic closed.

After some quick work, can someone check this for me? M1 = 1.2kg M2 = 3.2kg doing some math (that would be painful to type) yields: a = 4.45 m/s2 (a = acceleration of the blocks) Tr = 17.1 N Tc = 34.2 Thank you all for your help.

Sorry for not stating that it is a massless, frictionless pulley and cord. The only things with mass are the two blocks. I omitted their numbers so i can get a point in the right direction. There is a cord holding the pulley up (Tc) and then the tension of the rope (Tr) doing down, but twice...

so, would the tension of the cord holding up the pulley be twice the tension of the rope holding the masses?

Homework Statement A simple pulley has two masses on either end, one M1 and one M2. It is easy to find the acceleration of the two masses, and the tension on the rope suspending these masses. The pulley is suspended by a cord, however, and the question asks to find the tension in this cord...