For the situation shown in Figure 1, calculate: (a) FRFR size 12{F rSub { size 8{R} } } {} , the force exerted by the right hand, and (b) FLFL size 12{F rSub { size 8{L} } } {}, the force exerted by the left hand. The hands are 0.900 m apart, and the cg of the pole is 0.600 m from the left hand.

*Strategy*

Figure 1 includes a free body diagram for the pole, the system of interest. There is not enough information to use the first condition for equilibrium (netF=0(netF=0 size 12{F=0} {}), since two of the three forces are unknown and the hand forces cannot be assumed to be equal in this case. There is enough information to use the second condition for equilibrium (netτ=0)(netτ=0) size 12{ ital "net"`τ=0} {} if the pivot point is chosen to be at either hand, thereby making the torque from that hand zero. We choose to locate the pivot at the left hand in this part of the problem, to eliminate the torque from the left hand.

*Solution for (a)*

There are now only two nonzero torques, those from the gravitational force (τwτw size 12{τ rSub { size 8{W} } } {}) and from the push or pull of the right hand (τRτR size 12{τ rSub { size 8{R} } } {}). Stating the second condition in terms of clockwise and counterclockwise torques,

net τcw=–netτccw.net τcw=–netτccw. size 12{"net "τ rSub { size 8{"cw"} } ="net"τ rSub { size 8{"ccw"} } } {}

(1)or the algebraic sum of the torques is zero.

Here this is

τ
R
=
–τ
w
τ
R
=
–τ
w

(2)since the weight of the pole creates a counterclockwise torque and the right hand counters with a clockwise toque. Using the definition of torque, τ=rF
sin
θτ=rF
sin
θ size 12{τ= ital "rF""sin"θ} {}, noting that θ= 90ºθ= 90º size 12{θ} {} , and substituting known values, we obtain

0.900 mFR=0.600 mmg.0.900 mFR=0.600 mmg. size 12{ left (0 "." "900"" m" right ) left (F rSub { size 8{R} } right )= left (0 "." "600"" m" right ) left ( ital "mg" right )} {}

(3)Thus,

FR
=
0.6675.00 kg9.80 m/s2
=
32.7 N.
FR
=
0.6675.00 kg9.80 m/s2
=
32.7 N.

(4)*Solution for (b)*

The first condition for equilibrium is based on the free body diagram in the figure. This implies that by Newton’s second law:

F
L
+
F
R
–
mg
=
0
F
L
+
F
R
–
mg
=
0
size 12{F rSub { size 8{L} } +F rSub { size 8{R} } - ital "mg"=0} {}

(5)From this we can conclude:

F
L
+
F
R
=
w
=
mg
F
L
+
F
R
=
w
=
mg
size 12{F rSub { size 8{L} } +F rSub { size 8{R} } =w= ital "mg"} {}

(6)Solving for FLFL size 12{F rSub { size 8{L} } } {}, we obtain

F
L
=
mg
−
F
R
=
mg
−
32
.
7 N
=
5.00 kg
9.80
m/s
2
−
32.7 N
=
16.3 N
F
L
=
mg
−
F
R
=
mg
−
32
.
7 N
=
5.00 kg
9.80
m/s
2
−
32.7 N
=
16.3 N
alignl { stack {
size 12{F rSub { size 8{L} } = ital "mg" - F rSub { size 8{R} } } {} #
= ital "mg" - "32" "." 7 {} #
=0 "." "333" ital "mg" {} #
= left (0 "." "333" right ) left (5 "." "00"" kg" right ) left (9 "." "80"" m/s" rSup { size 8{2} } right ) {} #
="16" "." 3" N" {}
} } {}

(7)*Discussion*

**FLFL size 12{F rSub { size 8{L} } } {}** is seen to be exactly half of FRFR size 12{F rSub { size 8{R} } } {}, as we might have guessed, since FLFL is applied twice as far from the cg as FRFR.

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