Question

problem 11
watch the following video, then answer the question(s)
part a
if cord ab can withstand a maximum force of 3000 n, determine the force in bc so that the 200 kg crate can be supported.
1962 n 2990 n 2270 n 3000 n

Explanation:

Step1: Calculate the weight of the crate

The weight of the crate $W = mg$, where $m = 200\ kg$ and $g=9.81\ m/s^{2}$. So $W=200\times9.81 = 1962\ N$.

Step2: Set up force - equilibrium equations at point B

Let the force in $AB$ be $F_{AB}$ and the force in $BC$ be $F_{BC}$. Assume the system is in static - equilibrium. At point B, the sum of horizontal and vertical forces is zero. $\sum F_x = 0$ and $\sum F_y = 0$.
We know that $F_{AB}$ has a maximum value of $3000\ N$. Let's assume the angle between $AB$ and the horizontal is $\theta$. From the geometry of the problem (not shown in detail here, but assuming a right - angled situation for simplicity in a typical statics problem), if we consider the vertical component of $F_{AB}$ and the force in $BC$ and the weight of the crate.
Let's assume the vertical component of $F_{AB}$ is $F_{AB_y}$ and the horizontal component is $F_{AB_x}$.
If we consider the vertical equilibrium $\sum F_y=0$, we have $F_{AB_y}+F_{BC}-W = 0$.
If we assume the angle between $AB$ and the horizontal is such that we use the maximum value of $F_{AB} = 3000\ N$. Let's assume for a simple case where the crate is hanging vertically and $AB$ is at an angle to support it. If we consider the vertical equilibrium of forces at point B, and assume the vertical component of $F_{AB}$ is used to support part of the weight of the crate.
Let's assume the crate is in static equilibrium and we consider the forces acting on point B. The weight of the crate $W$ acts downwards.
We know that $W = 1962\ N$ and $F_{AB}\leq3000\ N$. In the vertical direction, if we assume the crate is suspended and the forces are in equilibrium, and if we assume that the vertical component of $F_{AB}$ is sufficient to support the crate, we note that the weight of the crate is $1962\ N$. Since the maximum force in $AB$ is $3000\ N$ and the weight of the crate is $1962\ N$, and considering the vertical equilibrium of forces at point B ($F_{AB_y}+F_{BC}-W = 0$), when the maximum force in $AB$ is not fully utilized in the vertical direction to support the crate, and in the most basic case of vertical equilibrium, the force in $BC$ is equal to the weight of the crate because the maximum capacity of $AB$ is not needed to support the weight in a simple static - equilibrium situation. So $F_{BC}=1962\ N$.

Answer:

$1962\ N$