A beam resting on two pivots has a length of L= 6.00 m and mass M=94.0kg. The pivot under the left end exerts a normal force n₁ on the beam, and the second pivot placed a distance = 4.00 m from the left end exerts a normal force n₂.

A woman of mass m=59.0kg steps onto the left end of the beam and begins walking to the right as in the figure below. The goal is to find the woman's position when the beam begins to tip.A woman of massmwalking across a beam which is resting on two pivots. The beam is of lengthLand massMand the woman is a distancexfrom the left end of the beam. The first pivot is directly under the left end of the beam and the second pivot is a distancefrom the first pivot at a shorter distance than the length of the beam.

(b) Where is the woman when the normal force

n₁

is the greatest? x= ___________________m (c) What is

n₁

when the beam is about to tip? ______N (d) Use the force equation of equilibrium to find the value of

n₂

when the beam is about to tip. ______N (e) Using the result of part (c) and the torque equilibrium equation, with torques computed around the second pivot point, find the woman's position when the beam is about to tip. x= _______________ m (f) Check the answer to part (e) by computing torques around the first pivot point. x= ________________ m (g)Except for possible slight differences due to rounding, is the answer the same? Yes or No _______