Show all work. Competitors in a tug-of-war (at least successful ones) lean back

Question

Show all work.

Competitors in a tug-of-war (at least successful ones) lean back as they pull. The diagram shows a competitor (weight W) in a tug-of-war. At the moment shown, the person is in equilibrium. The dot shows the position of the person's center of mass. Li is the distance from the person's foot to the center of mass; L2 is the distance from the foot to line of the rope. The rope is horizontal. The angle between the person's body and the horizontal is theta. Derive an equation relating the lean angle theta to the tension in the rope T. At a certain tension T the angle theta = 60 degree. If the tension is doubled, what will the new angle be? If the coefficient of static friction between the person's shoes and the ground is what will be the minimum angle 0 before the person starts to slip? Estimate a numerical value. Does your result seem reasonable?

Explanation / Answer

a) F*cos(theta) = T

F*sin(theta) = m*g

T*L2*sin(theta) = W*L1*cos(theta)

tan(theta) = W*L1/T*L2 = (W/T)*(L1/L2)

theta = tan^-1(W*L1/T*L2)

b) tan60 / tan(theta) = T/2T

tan(theta) = 2*tan60

theta = 73.88 degrees

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