The force required to deflect a rope under tensions?

Okay, since the rope is assumed to be massless and doesn't enter into the problem anywhere, then the tension on the rope is the same everywhere along it's length, regardless of the magnitude of the deflection: namely, 100 newtons.

The amount of force needed to make the deflection illustrated is 89.4427 newtons.

The rope makes an angle of 26.565 degrees with the vertical.

atan(.5)=26.565 degrees.

The sine of which is 0.44721

Multiplying this by 100 newtons gives us the horizontal component of the tension of EACH angled segment of rope.

Multiply by 2 to get the total: 89.4427 newtons.

That lower bend in the rope isn't realistic. Ropes don't curve. (Well they do under their own weight, but we're modeling this rope's weight as small compared to the weight below). So the rope will go from the hook to the pulley and then straight down. Draw a force diagram on the point at which the pulley grabs the rope.

You have the a tension of 100N pulling down. You have 100N of tension pulling up at an angle of arcsine of 0.5/2. That assumes that 2 is the length of rope between the pulley and the hook. If 2 is the height between the pulley and the hook, it would be arctangent of 0.5/2. Then you've got the force on the pulley (up and to the other side). The three forces have to sum up to zero.

so pulley's vertical force = T - T cos theta

pulley's horizontal force = T sin theta

Pulley's total force = T sqrt ((1 - cos theta)^2 + (sin theta)^2)

= T sqrt (2 - 2 cos theta)

So T never changes, put the fishscale will read zero when the weight hangs down (theta = 0) and sqrt(2) times the weight when it is pulled up to the same height as the hook.

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