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
You might be interested in
Answer:
-54.12 V
Explanation:
The work done by this force is equal to the difference between the final value and the initial value of the energy. Since the charge starts from the rest its initial kinetic energy is zero.
The change in electrostatic potential energy , of one point charge q is defined as the product of the charge and the potential difference.