A 4.0 m steel beam with a cross sectional area of 1.0 x 10 –2 m 2 and a Young\
ID: 2111591 • Letter: A
Question
A 4.0 m steel beam with a cross sectional area of 1.0 x 10–2 m2 and a Young's modulus of 2.0 x 1011 N/m2 is wedged horizontally between two vertical walls. In order to wedge the beam, it is compressed by 0.020 mm. If the coefficient of static friction between the beam and the walls is 0.70, what is the maximum weight can the beam hold without slipping?
A 4.0 m steel beam with a cross sectional area of 1.0 x 10–2 m2 and a Young's modulus of 2.0 x 1011 N/m2 is wedged horizontally between two vertical walls. In order to wedge the beam, it is compressed by 0.020 mm. If the coefficient of static friction between the beam and the walls is 0.70, what is the maximum weight can the beam hold without slipping?
Explanation / Answer
young's modulus ,Y=mg/(area*(delta length/ length))
2*10^11 = mg/(1*10^-2 * 0.02*10^-3 /4)
mg= 1*10^4
maximum weight = 0.7*mg = 7*10^3 N
=7 kN
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