To be able to solve for unknown forces and moments in rigid-body problems using
ID: 2997521 • Letter: T
Question
To be able to solve for unknown forces and moments in rigid-body problems using the equations of equilibrium.
For a rigid body to be in equilibrium, both the sum of the forces and the sum of the moments about an arbitrary point O must be zero:
?F=0
?MO=0
When all of the forces lie in the xy plane, the forces can be resolved into their x and y components, which results in the equations of equilibrium in two dimensions:
?Fx=0
?Fy=0
?MO=0
As shown, a uniform beam that weighs 66.0lb is attached to a wall at point A. The beam is subjected to three forces, F1 = 42.0lb , F2 = 12.0lb , and F3 = 27.0lb in the figure. The line of action of F2 passes through point A. If the wall can sustain a maximum moment of 585lb?ft about point A, what is the largest value for d, the beam's length, that preserves static equilibrium? The beam's width is negligible.
To be able to solve for unknown forces and moments in rigid-body problems using the equations of equilibrium. For a rigid body to be in equilibrium, both the sum of the forces and the sum of the moments about an arbitrary point O must be zero: ?F=0 ?MO=0 When all of the forces lie in the x½y plane, the forces can be resolved into their x and y components, which results in the equations of equilibrium in two dimensions: ?Fx=0 ?Fy=0 ?MO=0 As shown, a uniform beam that weighs 66.0lb is attached to a wall at point A. The beam is subjected to three forces, F1 = 42.0lb , F2 = 12.0lb , and F3 = 27.0lb in the figure. The line of action of F2 passes through point A. If the wall can sustain a maximum moment of 585lb?ft about point A, what is the largest value for d, the beam's length, that preserves static equilibrium? The beam's width is negligible.Explanation / Answer
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As shown, a uniform beam that weighs 63.0 lb is attached to a wall at point A. The beam is subjected to three forces, F1 = 50.0 lb, F2 = 16.0 lb, and F3 = 28.0 lb. (Figure Above) The line of action of F2 passes through point A. If the wall can sustain a maximum moment of 565 lboft about point A, what is the largest value for d, the beam's length, that preserves static equilibrium? The beam's width is negligible.
Note that the given weight vector of 63 lbs is not shown in the diagram, but that it acts downward at d/2.
The force F2 does NOT produce any moment about point A since its line of action passes thru A.
The three clockwise moments on the beam have to be balanced (equal to) the counterclockwise moment of 565 lb-ft at A .
Taking sum of moments about point A, we get: (63 lbs) (d/2) + (50 lbs) (4/5) (d/2) + (28 lbs) d = 565.
The factor of (4/5) is sine of the angle for force F1 .
Factor out d from the above moment eqn to get: d [ 31.5 + 20 + 28 ] = 565
Solve for d to get: d = 7.11 ft final answer.
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