Forces(5N, 030\(^o\)), (PN, 060\(^o\)), (QN, 150\(^o\)), (3N, 180\(^o\)) and (5N, 270\(^o\)) act on a body . If the system is in quilibrium, find, correct t...

We can solve this problem by applying the conditions for equilibrium:

1. ΣFx = 0 (Sum of forces in x-direction equals zero):
2. ΣFy = 0 (Sum of forces in y-direction equals zero):

Step 1: Resolve forces into components.

• 5 N, 30°:
  • Fx = 5 * cos(30°) ≈ 4.3 N (positive x-direction)
  • Fy = 5 * sin(30°) ≈ 2.5 N (positive y-direction)
• PN, 60°:
  • Fx = P * cos(60°) = P/2 (positive x-direction)
  • Fy = P * sin(60°) = P√3/2 (positive y-direction)
• QN, 150°:
  • Fx = -QN * cos(150°) = QN/2 (negative x-direction) (Since cosine is negative in this quadrant)
  • Fy = QN * sin(150°) = -QN√3/2 (negative y-direction)
• 3 N, 180°:
  • Fx = -3 * cos(180°) = -3 N (negative x-direction)
  • Fy = -3 * sin(180°) = 0 N (y-direction is zero)
• 5 N, 270°:
  • Fx = -5 * cos(270°) = 0 N (x-direction is zero)
  • Fy = -5 * sin(270°) = 5 N (negative y-direction)

Step 2: Apply equilibrium conditions.

ΣFx = 0:

4.3 + P/2 - QN/2 - 3 = 0

ΣFy = 0:

2.5 + P√3/2 - QN√3/2 - 5 = 0

Step 3: Solve the system of equations.

This is a system of two equations with two unknowns (P and Q). You can solve it using various methods like elimination or substitution. Here, we'll use elimination:

1. Multiply the top equation by -2√3: -8.6√3 - P√3 + Q√3 + 6√3 = 0

2. Add this equation to the bottom equation:

   -6.1√3 + Q√3 = 0

3. Divide both sides by √3:

   Q ≈ 6.1

4. Substitute this value of Q back into the top equation (we can choose either equation):

   4.3 + P/2 - (6.1)/2 - 3 = 0

5. Solve for P:

   P ≈ 8.8

Step 4: Round the answers to one decimal place.

Therefore, the values of P and Q are:

• P ≈ 8.8 N
• Q ≈ 6.1 N