1. Static Equilibrium Stress and strain
Discussions Apr 9
Static Equilibrium
Stress and strain

2. 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

3. 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

4. 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

5. 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 Extreme case is L’=L t2≤mg/2+Mg=400 N+1500 N=1900 N Watch out! t2,max=1800 N t1 P mg Mg

6. 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

7. Moving Day t2 t1’ P t1 mg -t1 Mg
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 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 P t1 mg
-t1 Mg

8. 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?

9. Moving Day-2 t2
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. P mg t1
t1’

10. 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 cross-sectional area of 4.0 cm2 and a compressive strength of 1.70x108 N/m2. Does his tibia break?

11. 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 cross-sectional 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)