What Is the Largest Mass the Ropes Can Support?

Answer:

170 kg

Explanation:

Costless-Body Diagram:

Let'southward offset past drawing a free trunk diagram for this trouble. Let's call the tension in the left rope $T_1$ and the tension in the right rope $T_2$ .

Now we tin can construct a free body diagram using the horizontal and vertical components of the 2 tension forces.

Allow'due south start with the left tension force:

The right tension force:

In lodge to see what these forces look similar on the diagram, look at the first image attached. (1)

At present, nosotros tin can create the FBD using the components of the tension forces and the force of gravity pulling on the crate, known as w.

The horizontal components of $T_1$ and $T_2$ go on the left and right of the FBD, respectively. The 2 vertical components go along pinnacle, and the weight force is on the bottom. We know that weight = mg.

To see what the free body diagram should wait like, see the second image attached. (two)

At present that we take the FBD, nosotros can write a system of equations using the sum of forces in both the x- and y-direction.

Since the crate is not moving in either the horizontal or vertical direction, we can say that the net force acting on the crate is 0.

$F_\text{net} = 0$

This ways that the sum of ten forces and the sum of y forces are both 0. Let'south set the top and right to be the positive direction and the bottom and left to be the negative direction. Now we tin write the system of equations:

There are many ways to solve this system of equations, just I will exist showing the tiptop 2 ways that I tend to solve these types of issues.

Method 1:

Solve for $T_1 \cdot sin(30)$ and $T_1 \cdot cos(30)$ in both equations.

We know that $\frac{sin\theta}{cos\theta}$ = $tan\theta$ , and so we can reorder both equations and divide them.

$tan(30)=\frac{mg-T_2\cdot sin(45)}{T_2\cdot cos(45)}$

Note that $\frac{T_1}{T_1}$ cancels, leaving u.s.a. with tan(thirty).

Now that we have this equation, nosotros can plug in known values. We are trying to solve for k, the largest mass of the crate that the ropes can back up. We know that:

Let'due south substitute these values into our equation:

$tan(30)=\frac{m(9.8)-(1500)\cdot sin(45)}{(1500)\cdot cos(45)}$

Multiply 1500 * cos(45) to both sides of the equation.

$tan(30) \cdot (1500 \cdot cos(45)) =9.8m-(1500 \cdot sin(45))$

Add together 1500 * sin(45) to both sides of the equation.

$tan(30) \cdot (1500 \cdot cos(45)) + 1500 \cdot sin(45)) =9.8m$

Divide both sides of the equation by ix.eight.

$\frac{tan(30) \cdot (1500 \cdot cos(45)) + (1500 \cdot sin(45)) }{9.8}= m$

Simplify.

$\frac{1673.032607}{9.8}=m$

$m=170.717613$

The largest mass the ropes can support is around 170 kg (circular downwardly, not upwardly).

Method 2:

We can use the system of equations again.

This fourth dimension, let's practice a little guess and check. Using the first equation, plug in 1500 N for T2 and solve for T1.

$T_1 \cdot cos(30) = (1500) \cdot cos(45)$

Carve up both sides of the equation past cos(30).

$T_1=\frac{1500 \cdot cos(45)}{cos(30)}$

Simplify.

$T_1=1224.744871 \ \text{N}$

Now plug in 1500 N for T1 and solve for T2.

$(1500) \cdot cos(30) = T_2 \cdot cos(45)$

Divide both sides by cos(45).

Since the tension cannot exceed 1500 North without the rope breaking, nosotros must employ the beginning substitution, where $T_1=1224.744871$ and $T_2=1500 \ \text{N}$ .