force of 2.0 kN acts on a beam of negligible weight, as shown. Determine the ang
ID: 1774002 • Letter: F
Question
force of 2.0 kN acts on a beam of negligible weight, as shown. Determine the angle necessary for the beam to be in equilibrium.
wrong answers --> 11.05, 1.2, 17.52 (degrees)
remember, I think the angle dealing with x and y components regarding normal force is 58, or 180-(32+90).. not just 32.
no one has been able to get this.
Exercise 5.3.08 A force of 2.0 kN acts on a beam of negligible weight, as shown. Determine the angle necessary for the beam to be in equilibrium. 1.6 m 0.5 m Al IB 32° 2.0 kN the tolerance is +/-2%Explanation / Answer
Balancing moment about A,
0.5*2000 = 2.1*By
By = 1000/2.1 = 476 N
Ay = 2000 - 476 = 1524 N
Ax = Ay*tan 32 degree = 1524/tan 32 degree = 952.3
Bx =952 N
Angle alpha = arctan (By/Bx)
= arctan (476/952)
= 26.56 degree answer
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