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The guy wire BD exerts on the telephone pole AC a force P directed along BD. Knowing that P must have a 720-N component perpendicu- lar to the pole AC, determine (a) the magnitude of the force P, (b) its component along line AC.

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Answer:

Explanation:

  • From the diagram, tanФ = opp/Adj = 2.4m/7m
  • tanФ = 0.3429. Ф = arctan ( 0.3429 )
  • Therefore, Ф = 18.92°

Given force perpendicular to AC from the diagram = 720N

  • Resolving vertically; PsinФ = 720

a) P = 720/sin18.92° = 2220N = the magnitude of the force P

b) To get component along line AC;

  • resolving horizontally, component along AC = PcosФ
  • = 2220 x cos 18.92° = 2100N

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The magnitude of the force P and its component along line AC are respectively; A) P = 2.22 kN and B) P_y = 2.10 kN

How to solve forces in equilibrium?

From the guy wire diagram attached i have drawn a free body diagram to solve for both the magnitude of the force P and its component along line AC.

A) From the free body diagram attached, we can use proportion to get;

P/Pₓ = 37/12

where;

Pₓ is given as 720 and P is the magnitude of force P.

Thus;

P = 720(37/12)

P = 2.22 KN

B) Similar to A above we can find the component of the force P along line AC as;

P_y = (35/12) * Pₓ

P_y = (35/12) * 720

P_y = 2.10 KN

Read more about resolution of forces at; https://brainly.com/question/25329636

Ver imagen AFOKE88
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