ENGG 205 Statics and Mechanics Question

[ps123]

Homework Statement

Each of the three cables can safely carry a maximum tension force of 14.1kN. Based on this criterion, what is the largest safe mass of the suspended ceiling? (unit: kilograms, kg)

Homework Equations

Sum of moments = 0
sum of forces in the x direction = 0
sum of forces in the y direction = 0

The Attempt at a Solution

I tried to sum up the tensions and then divide by 9.81 but i don't think its that easy.

 

[KataKoniK]

I would approach this problem by first understanding the physical properties of the cables and the suspended ceiling. I would also consider any other external factors that may affect the maximum tension force of the cables, such as wind or vibrations.

Then, I would calculate the weight of the suspended ceiling by using the formula w = mg, where w is the weight, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2). I would also consider the distribution of weight across the three cables and ensure that the maximum tension force of each cable is not exceeded.

Next, I would use the given criterion of 14.1kN as the maximum tension force for each cable. Using the formula F = ma, where F is the tension force, m is the mass, and a is the acceleration, I would solve for the maximum mass that each cable can safely support. Then, I would take the minimum value of the three calculated masses as the largest safe mass for the suspended ceiling.

Finally, I would double-check my calculations and consider any other safety factors before providing an answer. It is important to always approach problems like this with caution and attention to detail, as safety is of utmost importance.

 

[piercas]

it is important to approach problems like this using a systematic and logical approach. In this case, we can use the principles of statics and mechanics to solve this problem.

First, we need to identify all the forces acting on the suspended ceiling. In this case, there are three cables, each capable of carrying a maximum tension force of 14.1kN. Therefore, the total maximum tension force that can be applied to the suspended ceiling is 3 times 14.1kN, which is 42.3kN.

Next, we need to consider the weight of the suspended ceiling itself. This weight will act downward and will need to be balanced by the tension forces in the cables. To determine the weight of the ceiling, we can use the equation W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity (9.81 m/s^2).

Now, we can set up an equilibrium equation to solve for the maximum safe mass of the suspended ceiling. This equation will involve the sum of all the forces acting on the ceiling in both the x and y directions. Since the ceiling is not moving, the sum of all the forces in both directions must be equal to zero.

Therefore, we can set up the following equations:

Sum of forces in the x direction = 0:
T1cosθ + T2cosθ + T3cosθ = 0

Sum of forces in the y direction = 0:
T1sinθ + T2sinθ + T3sinθ - W = 0

Where T1, T2, and T3 are the tension forces in each cable, and θ is the angle between the cable and the horizontal.

Solving these equations simultaneously, we can find the value of W, which represents the weight of the suspended ceiling. Then, we can use the equation W = mg to solve for the maximum safe mass of the ceiling.

In conclusion, the largest safe mass of the suspended ceiling can be calculated by considering the maximum tension forces of the cables and the weight of the ceiling itself. By setting up equilibrium equations and solving for the weight of the ceiling, we can determine the maximum safe mass.