Determine the magnitude of the component of F directed along the axis of AB. Set F = 520 N .
2 answers:
Answer:
Force AB = 466.2N
Force AC = 380.7N
Complete Question:
The vertical force F acts downward at A on the two membered frame. Determine the magnitude of the component of F directed along the axis of AB . Set F=520 N
Determine the magnitude of the component of F directed along the axis of AC.
Attached is the image of the question.
Explanation:
Given the force distribution in the image attached:
Let X represent force AB and Y represent force AC
F = 520N
Resolving the force distribution to the vertical and horizontal components.
For vertical component (x axis)
F = Xcos45 + Ycos60 .....1
For horizontal component ( y axis)
Xsin45 = Ysin60 .....2
From equation 2
Y = Xsin45/sin60 ....3
Substituting equation 3 into equation 1
F = Xcos45 + (Xsin45/sin60)cos60
F = X ( cos45 + (sin45/sin60)cos60)
X = F/ ( cos45 + (sin45/sin60)cos60)
Substituting F = 520N and solving for X
X = 520/ ( cos45 + (sin45/sin60)cos60)
X = 466.2N
From equation 3.
Y = Xsin45/sin60
Y = 466.2sin45/sin60
Y = 380.7N
Therefore,
Force AB = 466.2N
Force AC = 380.7N
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