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Gnesinka [82]
2 years ago
5

The vertical force f acts downward at A on the two membered frames. Determine the magnitude of the two components of F directed

along axes of AB and AC. Set F=500N
Physics
1 answer:
galben [10]2 years ago
4 0

Answer:

The magnitude will be "353.5 N". A further solution is given below.

Explanation:

The given values is:

F = 500 N

According to the question,

In ΔABC,

⇒ \angle BCA = (90-30)

⇒             =60^{\circ}

then,

⇒ \angle BAC=(180-45-60)

⇒             =75^{\circ}

Now,

The corresponding angle will be:

⇒ \angle FAC=60^{\circ}

⇒ \angle FAB=70+60

⇒             =135^{\circ}

Aspect of F across the AC arm will be:

= F\times cos(60)

On putting the values of F, we get

= 500\times (.5)

= 200 \ Newton

Component F along the AC (in magnitude) will be:

= F\times cos(135)

= 500\times (-.707)

= -353.5 \ N \

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86)
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