Question

5. A 40-kg ornamental star is suspended by two chains fastened to

horizontal beams at different heights with angles as shown.
Determine the tension in each chain.


The diagram is shown below; I also know we have to split up the triangles to make two right angle triangles but I’m not sure where to go from there

Answer 1

Answer:We can start by resolving the forces acting on the ornamental star in the vertical direction and in the horizontal direction. Since the star is stationary, these forces must balance each other out.

Let's call the tension in the higher chain T1 and the tension in the lower chain T2. We can use trigonometry to find the vertical and horizontal components of each tension force.

For T1:

The vertical component is T1cos(40°) and the horizontal component is T1sin(40°).

For T2:

The vertical component is T2cos(30°) and the horizontal component is T2sin(30°).

Now, resolving the forces in the vertical direction, we have:

T1cos(40°) + T2cos(30°) - mg = 0

where m is the mass of the star and g is the acceleration due to gravity.

Substituting the values:

T1cos(40°) + T2cos(30°) - (40 kg)(9.81 m/s^2) = 0

T1cos(40°) + T2cos(30°) = 392.4 N

Next, resolving the forces in the horizontal direction, we have:

T1sin(40°) = T2sin(30°)

Now we have two equations with two unknowns:

T1cos(40°) + T2cos(30°) = 392.4 N

T1sin(40°) = T2sin(30°)

Solving these equations simultaneously, we get:

T1 = 266.4 N

T2 = 205.1 N

Therefore, the tension in the higher chain (T1) is 266.4 N and the tension in the lower chain (T2) is 205.1 N.

Explanation: