Discussions Apr 9
Static Equilibrium
Stress and strain
• Q1: Two people stand at one
end of a uniform platform
supported by cables at each end.
In equilibrium, what do we expect
to find about the tensions in the
two cables?
A) The tensions are equal
B) The tension in the cable
farther from the people is larger
C) The tension in the cable
closer to the people is larger
D) More information is required
• Q1: Two people stand at one
end of a uniform platform
supported by cables at each end.
In equilibrium, what do we expect
to find about the tensions in the
two cables?
A) The tensions are equal
B) The tension in the cable
farther from the people is larger
C) The tension in the cable
closer to the people is larger
D) More information is required
High rise physics
P1: You have a summer job working
downtown washing windows on skyscrapers
(the pay is great and so are the medical
benefits). The platform you and your partner
are using to get to the windows is 1.5 meter
wide and 6.0 meters long. You know from
hauling the platform out of your truck
countless times that it has a mass of 80 kg.
The combined mass for you and your
partner is 150 kg. Suppose you know the
cables can safely support 180 kg each. Is it
safe for both of you to stand at the same end
of the platform?
Use g=10 m/s2
High rise physics
Net force on platform is zero:
Fnet=t1+t2-mg-Mg=0
mg=800 N Mg=1500 N
t1+t2=2300 N
t2
Net torque about Point P on platform is zero:
t=t2L-mgL/2-MgL’=0
t1
Extreme case is L’=L
t2≤mg/2+Mg=400 N+1500 N=1900 N
Watch out! t2,max=1800 N
P
mg
Mg
Moving Day
• P2: You are watching a some of your friends
preparing to raise a 300 kg piano from the
street up to the third floor window of an
apartment using a strong cable pulled by a
winch inside the building on the fourth floor.
They have set up a pulley on the end of a 100
kg steel beam that is 2 meter long, which is
bolted to the outer wall of the building. The
beam is supported by the same type of cable
that is attached 1.5 meter from the wall. The
cable forms a 30-degree angle with respect to
the beam. You inquire about the strength of
the cable and are told that it is capable of
supporting 1000 kg. What do you advise
them about the safety of what they are about
to do?
Use g=10 m/s2
Moving Day
t2
Look at torque on the beam:
t=t2(1.5 m) sin(30°)-t1(2.0 m)-mg(1.0 m) =0
Note that t1=Mg=3000 N
t 1’
P
t1
mg
t1’ is the same strength, but exerts a negligible torque
about P.
mg=1000 N
t2=[(3000 N)(2.0 m)+(1000 N)(1.0 m)]/[(1.5 m)•sin(30°)]
=[7000 N•m]/[0.5•1.5 m]=9333 N
This is less than t2,max=(1000 kg)g=10000 N
-t1
Mg
Moving Day-2
• Similar situation:
– 300 kg piano
– 100-kg steel beam that is 2.0 meter long,
which is firmly bolted to the outer wall of
the building.
– cable that is attached 1.5 meter from the
wall. The cable forms a 30-degree angle
with respect to the beam.
– Cables are capable of supporting 1000
kg.
• Q2: But, now suppose that the winch
is located at street level (but firmly
anchored): What’s different?
What would you advise them about
the safety in this situation?
Moving Day-2
• Suppose that the winch is located at
street level (but firmly anchored):
What’s different?
What would you advise them about
the safety in this situation?
• Now t1’ does not exert a negligible
torque. It is about the same as the
torque from t1. The cable providing
the supporting tension, t2, will break.
t2
P
mg
t 1’
t1
Hard Landing
• A 75-kg man unwisely jumps down from a height of
3.0 meter and lands on hard ground with stiff legs so
that his center of mass stops over a distance of
3.0 cm. The average force on his legs while
stopping is therefore approximately 100 times the
man’s weight.
P3: The brunt of this force is experienced by the
man’s tibia (lower leg bone), which has a crosssectional area of 4.0 cm2 and a compressive strength
of 1.70x108 N/m2. Does his tibia break?
Hard Landing
• A 75-kg man unwisely jumps down from a height of
3.0 meter and lands on hard ground with stiff legs so
that his center of mass stops over a distance of
3.0 cm. The average force on his legs while
stopping is therefore approximately 100 times the
man’s weight.
The force is 100*(75 kg)*g=75000 N
P3: The brunt of this force is experienced by the
man’s tibia (lower leg bone), which has a crosssectional area of 4.0 cm2 and a compressive strength
of 1.70x108 N/m2. Does his tibia break?
The compressive strength is
(1.70x108 N/m2)(4.0x10-4 m2)=68000 N. (Oh, Snap)