[jyl]

two masses hang from a connecting rope over a pulley
rope has no stretch, pulley has no friction, no air resistance
the larger mass is M, the smaller mass is m, M > m
when the masses are released, they both accelerate
mass M has acceleration - a and mass m has acceleration a
the difference in the signs is because mass M is accelerating down and mass m is accelerating up
but the absolute value of the accelerations is the same, a, because they are connected by the rope

the forces acting on mass M to produce acceleration - a are gravity (down) and the force being transmitted by the rope (up)
gravity force is M * - g the negative sign is because gravity is pointing down
rope force call it T for tension and it is positive because the rope tension is pointing up
their sum produces the net force on mass M which is M * - a
so M * - a = M * - g + T
so M * a = M * g - T
so T = M * g - M * a

the forces acting on mass m to produce acceleration a are also gravity and the force being transmitted by the rope
gravity force is m * - g the negative sign is because gravity is pointing down
rope force call it T for tension and it is positive because the rope tension is pointing up
their sum produces the net force on mass m which is m * a
so m * a = m * - g + T
so m * a = T - m * g
so T = m * a + m * g

Since T = T
M * g - M * a = m * a + m * g
( M + m ) * a = ( M - m ) * g
a = [ (M - m) / ( M + m ) ] * g

M = 350 lb, m = 175 lb, g = 32 ft / sec^2
so a = 10.6667 ft / sec^2

Now solve for T
T = m * a + m * g
T = 175 lb * 10.6667 ft / sec^2 + 175 lb * 32 ft / sec^2
= 7466.6667 ft * lb / sec^2

The force on the pulley, which we'll call P for pull, is 2 * - T
P = 2 * - T
= -14,933.3333 ft * lb / sec^2

The pulley is attached to a platform, which must support "X pounds". That means X pounds in Earth's gravity g, so

-14,933.333 ft * lb / sec^2 = X lb * 32 ft / sec^2
X = [ -14,933.333 ft * lb / sec^2 ] / [32 ft / sec^2 ]
= -466.6667 lb

So if the platform must support 467 lb, and for man-lifting stuff you want a 3X safety factor (I'm just guessing), the platform needs to be rated for
467 lb * 3 = 1,400 lb.

That ignores shock loading, like if the stuntmen allow any slack in the rope, and elasticity of the rope. Guys who use elastic ropes to yank stuck vehicles from mud know that these are a huge deal.

Sorry for all the tedious units, that's how I was always taught to do physics. Write out the units explicitly and carry them through the whole calculation, if you end up with a mass in lb / sec you know you screwed up.

By the way, you see from the above that the platform is not necessarily supporting the combined mass of M and m. If M = m, it is the case, X = M + m and neither mass is moving, a = 0. But if M > m, the X < M + m, and the platform supports less weight as M / m gets larger. When m = 0, a = g and T = 0, P = 0, X = 0.

Imagine an extreme case, and you can see it. Suppose M = 350 lb and m = 0.0625 lb, the two stuntmen are lifting just 1 ounce. The stuntmen free-fall as if they were not holding a rope ( a = g appx ) , the rope flies through the pulley with no resisting weight on the other side, and the pulley is supporting almost no weight (X = 0 appx )

So, if your stunt involves two 175 lb stuntmen pulling a midget, this might all be okay.