Find force?????

A 64 kg rock climber is climbing a "chimney" between two rock slabs. The static coefficient of friction between her shoes and the rock is 1.06; between her back and the rock it is 0.84. She has reduced her push against the rock until her back and her shoes are on the verge of slipping. What is her push against the rock?





What fraction of her weight is supported by the frictional force on her shoes?

Find force?????
the gravitational force pulls her down by


Fg=m*g


she must compensate (exactly) this force by the friction force. The friction force is given by the force acting on the rock and the static coefficient f


Ff=F*f





because of action and reaction, the force of her shoes Fs and her back Fb will be the same, only each of them has different coefficient of friction:


Fb=F*fb, Fs=F*fs


And Fb + Fs compensate the gravitational force:


Fg = Fb + Fs


mg=F(fb+fs)





from here her push against rock is


F=mg/(fb+fs)





putting values there


F=64*9.81/(1.06+0.84)=330.4N





hope it helps:-)



dental