TP 11.86

 

 

A 64.0-kg person stands on a weightless diving board supported by two pillars, one at the end of the board, the other 1.10-m away. The pillar at the end of the board exerts a downward force of 828 N. The board has a length of 3.00.

  1. How far from that pillar is the person standing?

tN2= 0

tN1 = + (828 N)(1.10 m) = 911 Nm

tMg = - (64.0 kg)(9.80 m/s2)(x)

St = 0

(64.0 kg)(9.80 m/s2)(x) + 911 Nm = 0          x = (911 Nm)/ (64.0 kg)(9.80 m/s2)

                                                                          = 1.45 m (from the second pillar or 2.55m from the first)

  1. Find the force exerted by the second pillar. 

 

(M g)x = M g cos (- 90o) = 0

(M g)y = M g sin (- 90o) = - (64kg) (9.80 m/s2) = - 627 N

 

(N1)x =  N1 cos (-90o) = 0

(N1)y =  N1 sin (-90o) = - 828 N

 

(N2)x =  N2 cos (90o) = 0

(N2)y =  N2 sin (90o) = N2

 

SFx = 0

SFy = 0                     - 828 N - 627 N + N2  = 0    N2 = 1455 kg