#354 Determine the tensions developed in wires

Determine the tensions developed in wires - Mechanical Engineering

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Free Chegg Question

Determine the tensions developed in wires CD, CB, and BA and the angle θ required for equilibrium of the 30 lb cylinder E and the 60 lb cylinder F.

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Free Chegg Answer 1

 Tension in CD = 65.88 N
Tension in CB = 57.13 N
Tension in BA = 80.69 N
θ = 2.96°
worked out as under.

Let tension in AB = x lb, in BC = y lb and in CD = z lb.

For equilibrium of F and E in the horizontal direction,
xcos45° = ycosθ = zcos30°
=> x/√2 = (√3/2) z ... ( 1 )

For equilibrium of F and E in the vertical direction,
60 - xsin45° = ysinθ = zsin30° - 30
=> x/√2 = 90 - z/2 ... ( 2 )

From (1) and (2),
(√3/2) z + z/2 = 90
=> z = 180/(√3 + 1) = 90(√3 - 1) = 65.88 N
Plugging in (1),
x = √(3/2) z = √(3/2) * (65.88) = 80.69 N

Next, squarring and adding,
ycosθ = zcos30° and ysinθ = zsin30° - 30
=> y^2 = z^2 - 30z + 900
=> y = √[(65.88)^2 - (30)*(65.88) + 900] = 57.13 N

Also,
ycosθ = zcos30°
=> cosθ = (z/y)cos30° = [(65.88)/(57.13)] cos30° = 0.9987
=> θ = 2.96°

 

Free Chegg Answer 2

Let tension in AB = x lb, in BC = y lb and in CD = z lb.

For equilibrium of F and E in the horizontal direction,
xcos45° = ycosθ = zcos30°
=> x/√2 = (√3/2) z ... ( 1 )

For equilibrium of F and E in the vertical direction,
60 - xsin45° = ysinθ = zsin30° - 30
=> x/√2 = 90 - z/2 ... ( 2 )

From (1) and (2),
(√3/2) z + z/2 = 90
=> z = 180/(√3 + 1) = 90(√3 - 1) = 65.88 N
Plugging in (1),
x = √(3/2) z = √(3/2) * (65.88) = 80.69 N

Next, squarring and adding,
ycosθ = zcos30° and ysinθ = zsin30° - 30
=> y^2 = z^2 - 30z + 900
=> y = √[(65.88)^2 - (30)*(65.88) + 900] = 57.13 N

Also,
ycosθ = zcos30°
=> cosθ = (z/y)cos30° = [(65.88)/(57.13)] cos30° = 0.9987
=> θ = 2.96°