Determine the magnitude of the component of f directed along the axis of ab. set f = 520 n .

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

1/0.14nm = 10^10/1.4 a lot