The drawing shows a circus clown who weighs 960 N.?

Question

The drawing shows a circus clown who weighs 960 N. The coefficient of static friction between the clown's feet and the ground is 0.45. He pulls vertically downward on a rope that passes around three pulleys and is tied around his feet. What is the minimum pulling force that the clown must exert to yank his feet out from under himself?
Drawing: http://images.google.com/imgres?imgurl=http://webassign.net/CJ/04_58.gif&imgrefurl=http://www.webassign.net/v4cgi/student.pl%3FWebAssignUsername%3Ddemo%26WebAssignInstitution%3Ddemo%26WebAssignPassword%3Ddemo%26Department%3Dv4demo%26course%3D11,1%26struct%3DABDiAYAxBxBvEAEvBMDFDMCXAkCfEnBUAeDUEIDBBcAEAaERCEATAkEjDqCaBGEFBoELArCQDHCmAtCXBFDRBhCxuUCPEPDiDcAbBgEMEmCjDtErCXDkDh&h=321&w=225&sz=12&hl=en&start=1&um=1&tbnid=k3sQw_VDwFG1qM:&tbnh=118&tbnw=83&prev=/images%3Fq%3Dphysics%2BThe%2Bdrawing%2Bshows%2Ba%2Bcircus%2Bclown%2Bwho%2Bweighs%2B.%2BThe%2Bcoefficient%2Bof%2Bstatic%2Bfriction%2Bbetween%2Bthe%2Bclown%2527s%2Bfeet%2Band%2Bthe%2Bground%2Bis%2B.%2BHe%2Bpulls%2Bvertically%2Bdownward%2Bon%2Ba%2Brope%2Bthat%2Bpasses%2Baround%2Bthree%2Bpulleys%2Band%2Bis%2Btied%2Baround%2Bhis%2Bfeet.%2BWhat%2Bis%2Bthe%2Bminimum%2Bpulling%2Bforce%2Bthat%2Bthe%2Bclown%2Bmust%2Bexert%2Bto%2Byank%2Bhis%2Bfeet%2Bout%2Bfrom%2Bunder%2Bhimself%253F%26svnum%3D10%26um%3D1%26hl%3Den%26safe%3Doff%26rls%3DDKUS,DKUS:2006-32,DKUS:en%26sa%3DN

Answer 1

When the clown pulls on the rope, he takes some weight off his feet. So that affects the friction equation. Check and see if you agree this equation applies to the situation.

Ff = mu*(W-Ff)

When the sideways pull on his feet equals the friction between the feet and ground, he'll flip. The pulleys just change the direction of the pull he gives the rope (no mechanical advantage in this setup). So if you evaluate the equation, the resulting Ff should be the required minimum.

Answer 2

Your drawing is lacking yet while memory serves me precise, I rather have solved a similar undertaking earlier and that i might go with to assume that the prognosis for this undertaking is the comparable. enable F = pulling stress for the clown to be yanked out from his ft f = frictional stress between the floor and the clown's ft and so, the equation for that's F - f = 0 observe that f = mu(frequent stress) the place mu = coefficient of friction = 0.537 (given) frequent stress = weight of the clown = 792 N (given) hence, substituting suitable values, F - (0.537)(792) = 0 F = 0.537*792 F = 425.304 N desire this facilitates.