A 4.0-m long steel beam with a cross-sectional area of 1.0 times 10^-2 m^2 and a

Question

A 4.0-m long steel beam with a cross-sectional area of 1.0 times 10^-2 m^2 and a Young's modulus 2.0 times 10^11 N/m^2 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 the maximum mass (including its own) it can bear without slipping is: 0 3.6kg 36 kg 71 kg 710kg Two supports, made of the same material and initially of equal length, are 2.0 m apart. A stiff board with a length of 4.0m and a mass of 10 kg is placed on the supports, with one support at the left end and the other at the midpoint. A block is placed on the board a distance of 0.50 m from the left end. As a result the board is horizontal. The mass of the block is: zero 2.3 kg 6.6 kg 10 kg 20 kg

Explanation / Answer

47)

Stress = e*E

Stress = F/A

F/A = e*E
F = A*e*E

Li = 4 m
Lf = 4 m - 0.020 mm = 4 m - .00002 m = 3.99998 m
e = (Li - Lf)/Li = .00002 m / 4 m = 0.000005

F = 1.0*10^-2 m^2 * 0.000005 * 2.0*10^11 N/m^2 = 10000 N

friction = u*F

Sum of the vertical forces = 0 = m*g - friction
m = friction / g = u*F/g

Plug in numbers
m = 0.70 * 10000 N / 9.81 m/s^2 = 710 kg

48)

By summing vertical forces to zero

2F - (M + 10)g = 0
F = (M + 10)g/2

By summing the moments about the right end of the board to zero, we will able to find the unknown mass M
Assume a clockwise moment is positive

F[4] + F[2] - Mg[3.5] - 10g[2] = 0
F[6] - Mg[3.5] - 10g[2] = 0
(M + 10)g/2[6] - Mg[3.5] - 10g[2] = 0
(M + 10)g[3] - Mg[3.5] - 10g[2] = 0
(M + 10)[3] - M[3.5] - 10[2] = 0
3M + 30 - 3.5M - 20 = 0
10 = 0.5M
M = 20 kg

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