Q1:In a materials testing laboratory, a metal wire made from a new alloy is foun
ID: 1490155 • Letter: Q
Question
Q1:In a materials testing laboratory, a metal wire made from a new alloy is found to break when a tensile force of 86.2 N is applied perpendicular to each end.
a.f the diameter of the wire is 1.01 mm , what is the breaking stress of the alloy?
Q:2A circular steel wire 2.10 m long must stretch no more than 0.30 cm when a tensile force of 360 Nis applied to each end of the wire.
a.What minimum diameter is required for the wire?
b.Express your answer using two significant figures.
q:3A uniform rod is 2.40 m long and has mass 1.60 kg . A 2.20 kg clamp is attached to the rod.
a.How far should the center of gravity of the clamp be from the left-hand end of the rod in order for the center of gravity of the composite object to be 1.25 m from the left-hand end of the rod?
Express your answer with the appropriate units.
Explanation / Answer
Q1: Answer
Ftu = F/A = 86.2/[(/4)*.00101²] = 107.59 MPa
Q2. Answer
Young's modulus of steel wire = Y's = (F*L)/(l*A)
where F is the force, L and A initial length and cross sectional area of the wire respectively and l is the actual stretch due to F,
Hence A = (F*L)/(l*Y's) = (360*2.10)/[3*(10^-3)*2*10^11] = 1.26mm
Q3. Answer
(1.60kg/2.4m) = 0.667kg/m. mass of rod.
(1.25m x 0.667) = mass left of 0.83375kg.
Mass right = (1.60 - 0.83375) = 0.76625kg.
Torque left = (1.25/2) x 0.83375 = 0.52109kg/m.
Torque right = 1/2 (2.4 - 1.25) x 0.76625 = 0.44059375kg/m.
Difference = ( 0.52109 - 0.44059375) = 0.0805kg/m.
1.25 + ( 0.0805/2.2) = 0.37m. from the left.
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