Equilibrium and force

Question

When a sail is hoisted, the rope or halyard passes over a pulley and down the mast, where it is often wrapped around a cleat at the bottom. to effect the final tightening the halyard is grasped at the middle, if possible, and pulled away from the mast, after which the slack so gained is taken up at the cleat. If the halyard is 20 m from pulley to cleat and is pulled at the middle, so that it is 1.0 m out from the mast, with a force of 20 N, what is the tension in the halyard?

Answer 1

So Leah Anne F.,

Draw your object and start labeling force vectors. The tension in the halyard is decomposable into dual vertical components and dual corresponding horizontal components. Each triangle: middle of mast : middle of halyard : top (or bottom) of halyard therefore has the horizontal component as the tangent function of the subtended angle and the tension in the halyard. So, for example, 10N/T = (1/10) (You don't need to calculate the tangent of the subtended angle, since the sides are given to you already)

Note two things: 1) you need to be clear on the legs of the triangle formed! and 2) you need to remember that each side of the halyard (i.e., above and below your pull-point, respectively) are providing tension forces which are summable at the pull point. Since there are two such forces, each one only needs to provide half of the lateral pull force, thus 10N.

Written that way, it's trivial to see that T=100N

Note also that for small halyard deflections, the amount of stretch in the line is small, but that may not be the case at larger deflections -- thus, tension may not be an exact cosecant function of deflection.

-- Cheers, --Mr. d.