IGCSE/GCSE/O & A Level/IB/University Student Forum
Qualification => Subject Doubts => GCE AS & A2 Level => Sciences => Topic started by: helllife on February 18, 2010, 12:17:14 pm
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I have some questions on elasticity.....
1) A lift of mass 1000kg hangs from the ends of steel cable.The maximum upward acceleration of the lift is 1.2m/s2.If the maximum safe stress for the steel is 1.1x10^8 Pa, calculate the minimum diameter of the cable.
2) A steel wire and a brass wire of the same cross sectional area of 1 mm2 and length 4mm are suspended from the same point. The lower ends of the wire are joined together.WHAT is the common extension of the wires when the mass 10.00 kg end of the composite wire?( young moduus for steel and brass are2.00x1011Pa and 9.00x1010Pa respectively.)
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1 hour
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1)
m=1000kg
a=1.2m/s2
Stress=1.1X10^8
F=ma
=1000X1.2=1200N
Stress=F/A
A=1200/1.1X10^8
=1.1X10^-5
A=[pir^2
1.1X10^-5/pi=r^2
r=1.86X10^-3
d=3.7X10^-3m
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2)
A=1mm2
L=4mm
Youngs modulus=( F/A)/( delta L/L)
=FL/ AdeltaL
For steel
Delta L=FL/A X Young modulus(Y)
=100X0.004/ 1X10^-6 X 2X10^11
Note you have to convert to standard units
4mm=0.004 m
1mm2=0.001X0.001=1X10^-6m2
delta L=2X10^-6m
For Bass
Repeat same way
delta L=4.4X10^-7
Total extension= sum of two changes in length=2.44X10^-6m
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Thanks nid. The database is being updated.
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Thanks nid. The database is being updated.
Don't mention it... :)