Fright
Diving board
has negligible mass
4.5 m
1.5 m
Fleft
580 N
since parts a&b ask about the left pedestal, we should choose the location of the right pedestal as the point from which to sum the torques. This will effectively eliminate that force from our equation. no friction
Sample problem
A 500. N person is 4.0 meters up a 200. N , 10.0 m long ladder which is positioned against a frictionless wall with its base 3.0 m out from the wall
Find: Force from wall
Normal Force at floor
force of Static friction with Floor
3.0 m
FBD
Process
1) draw neat labelled FBD
2) write a τnet = 0 equation
3) Solve #2 for unknown and find other values using Fnet x = 0 and Fnet y = 0
40°
A
70.0°
B
200 N
Find Tensions in strings A & B
Since, all the forces involved here act "through"
a common point (see FBD below), we don't use
torque at all. We just use Fnet = 0.
Since this is two dimensional, we can break
Fnet = 0
Fnet x = 0
TA
TB
Fnet y = 0
from Fnet x = 0
we can see that
TBX ­ TAX = 0
TA
TAX
TB
TBX
200 N
200 N
from FNET y = 0
we can see that the sum to
of the two upward components (TAY + TBY) should balance the 200 N weight
TA
TAY
TBY
TB
200 N
If this make sense, the rest of this is just a matter of doing math and SOHCAHTOA stuff