A bag of cement whose weight is Fg hangs in equilibrium from...

A bag of cement whose weight is $F_{g}$ hangs in equilibrium from three wires as shown in Figure P5.24. Two of the wires make angles $\theta_{1}$ and $\theta_{2}$ with the horizontal. Assuming the system is in equilibrium, show that the tension in the left-hand wire is $$ T_{1}=\frac{F_{g} \cos \theta_{2}}{\sin \left(\theta_{1}+\theta_{2}\right)} $$

A bag of cement whose weight is $F_{g}$ hangs in equilibrium from three wires as shown in Figure
P5.24. Two of the wires make angles $\theta_{1}$ and $\theta_{2}$ with the horizontal. Assuming the system is in equilibrium, show that the tension in the left-hand wire is
$$
T_{1}=\frac{F_{g} \cos \theta_{2}}{\sin \left(\theta_{1}+\theta_{2}\right)}
$$

Key Concepts

Trigonometric Relationships

Trigonometric functions, particularly sine and cosine, are used to relate the angles the wires make with the horizontal to their force components. This relationship is key to deriving the expression for the tension in the wires.

Force Balance Equations

Setting up equations for the horizontal and vertical force balances is a fundamental approach in statics. These equations, derived from the equilibrium conditions, allow for solving the unknown tensions in the wires in terms of the given weight and angles.

Mechanical Equilibrium

Mechanical equilibrium in statics means that the sum of all forces and moments acting on a system is zero. In this problem, the bag of cement is stationary, thus the vector sum of all tension forces and the gravitational force must cancel out.

Force Decomposition

Resolving forces into horizontal and vertical components is essential when forces act at angles. By using sine and cosine, one can analyze the contributions of each force direction separately, which is critical to setting up the correct equilibrium equations.

Transcript

00:01 Now in this question we have a cement with weight fg being hung over the ceiling and we want to prove that the tension t1 over here can be written in terms of this expression.

00:20 So we got to start by writing down a few newton's law, newton's second law equations because the entire system is in equilibrium.

00:31 So there's no net force in both the x and the y direction.

00:39 Now if you consider this point over here, this center point.

00:46 What are the forces acting over here? so over here we have t3, pointing down, we have t2, this angle and also t1.

01:02 So we can actually apply the second level.

01:05 Law over here in both the x and the y direction this is the y direction and this is the x direction so we're going to start with x first so in the x direction we have positive x component coming from t2 and we have negative x component from t1 so in order to have net force to be zero it means that they must both be equal in magnitude the x component so for the x component of t1 we need to know what is this angle this angle over here is theta 1 right by the opposite angle right because over here this is theta 1 same thing with t 2 this will be theta 2 so the x components are actually quite easy to find it's just t1 cosine theta 1 as he goes to t 2 2 .0 .2 .2 .2 .2 .2.

02:18 Now we consider the y components.

02:24 So for the y components, the positive y would come from both t1 and t2, the negative would come from t3.

02:33 But t3 is just equals to the weights of our cement, such fg.

02:43 So fg must be equal in magnitude to the sum of the y components from t1 and t2.

02:49 T1, sine t31, and t2, sine t32...

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