Question

1.
X, Y and Z are the weights of three objects
suspended by pulleys as shown.
Assuming the pulleys in this system are frictionless
and weightless and that Z= 12 N, what are the
values of X and Y?
24°
50°
Х
Y
Z

Answer 1

Answer:

The weight of the object X is approximately 3.262 N (Acting downwards)

The weight of the object Y is approximately 8.733 N (Acting downwards)

Explanation:

The question can be answered based on the principle of equilibrium of forces  

The given parameters are;

The weight of Z = 12 N (Acting downwards)

The weight of the pulleys = Negligible

From the diagram;

The tension in the in the string attached to object Z = The weight of object Z = 12 N

The tension in the in the string attached to object X = The weight of the object X  

The tension in the in the string attached to object Y = The weight of the object Y

Given that the forces are in equilibrium, we have;

The sum of vertical forces acting at a point, $\Sigma F_y$ = 0

Therefore;

$T_{1y} + T_{2y} + T_{3y} = 0$

$T_{1y} = -( T_{2y} + T_{3y} )$

Where;

$T_{1y}$  = The weight of object Z = 12 N

$T_{1y}$  = 12 N

$T_{2y}$ = The vertical component of tension, T₂ = T₂ × sin(24°)

∴  $T_{2y}$ = T₂ × sin(156°)

Similarly;

$T_{3y}$ = T₃ × sin(50°)

From $T_{1y} = -( T_{2y} + T_{3y} )$, and $T_{1y}$  = 12 N, we have;

12 N = -(T₂ × sin(156°) + T₃ × sin(50°))...(1)

Given that the forces are in equilibrium, we also have that the sum of vertical forces acting at a point, ∑Fₓ = 0

Therefore at point B, we have;

T₁ₓ + T₂ₓ + T₃ₓ = 0

The tension force, T₁, only has a vertical component, therefore;

∴ T₁ₓ = 0

∴ T₂ₓ + T₃ₓ = 0

T₂ₓ = -T₃ₓ

T₂ₓ =  T₂ × cos(156°)

T₃ₓ = T₃ × cos(50°)

From T₂ₓ = -T₃ₓ, we have;

T₂ × cos(156°) = - T₃ × cos(50°)...(2)

Making T₃ the subject of equation (1) and (2) gives;

Making T₃ the subject of equation in equation (1), we get;

12 = -(T₂ × sin(156°) + T₃ × sin(50°))

∴ T₃ = (-12 - T₂ × sin(156°))/(sin(50°))

Making T₃ the subject of equation in equation (2), we get;

T₂ × cos(156°) = - T₃ × cos(50°)

∴ T₃ = T₂ × cos(156°)/(-cos(50°))

Equating both values of T₃ gives;

(-12 - T₂ × sin(156°))/(sin(50°)) = T₂ × cos(156°)/(-cos(50°))

-12/(sin(50°)) = T₂ × cos(156°)/(-cos(50°)) + T₂ × sin(156°)/(sin(50°))

∴ T₂ = -12/(sin(50°))/((cos(156°)/(-cos(50°)) + sin(156°)/(sin(50°))) ≈ -8.02429905283

∴ T₂ ≈ -8.02 N

From T₃ = T₂ × cos(156°)/(-cos(50°)), we have;

T₃ = -8.02× cos(156°)/(-cos(50°)) = -11.3982199717

∴ T₃ ≈ -11.4 N

The weight of the object X = -T₂ × sin(156°)

∴ The weight of the object X ≈ -(-8.02 × sin(156°)) = 3.262 N

The weight of the object X ≈ 3.262 N (Acting downwards)

The weight of the object Y = -(T₃ × sin(50°))

∴ The weight of the object Y = -(-11.4 × sin(50°)) ≈ 8.733 N

The weight of the object Y ≈ 8.733 N (Acting downwards)