Answer:
170 kg
Explanation:
Free-Body Diagram:
Let's start by drawing a free body diagram for this problem. Let's call the tension in the left rope [tex]T_1[/tex] and the tension in the right rope [tex]T_2[/tex].
Now we can construct a free body diagram using the horizontal and vertical components of the two tension forces.
Let's start with the left tension force:
- Horizontal component (x): [tex]T_1 \cdot cos(30)[/tex]
- Vertical component (y): [tex]T_1 \cdot sin(30)[/tex]
The right tension force:
- Horizontal component: [tex]T_2 \cdot cos(45)[/tex]
- Vertical component: [tex]T_1 \cdot sin(45)[/tex]
In order to see what these forces look like on the diagram, look at the first image attached. (1)
Now, we 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 [tex]T_1[/tex] and [tex]T_2[/tex] go on the left and right of the FBD, respectively. The two vertical components go on top, and the weight force is on the bottom. We know that weight = mg.
To see what the free body diagram should look like, see the second image attached. (2)
Now that we have the FBD, we 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.
- [tex]F_\text{net} = 0[/tex]
This means that the sum of x forces and the sum of y forces are both 0. Let's set the top and right to be the positive direction and the bottom and left to be the negative direction. Now we can write the system of equations:
- [tex]\sum F_x = T_2 \cdot cos(45) - T_1\cdot cos(30) = 0[/tex]
- [tex]\sum F_y = T_1 \cdot sin(30) +T_2\cdot sin(45) - mg = 0[/tex]
There are many ways to solve this system of equations, but I will be showing the top two ways that I tend to solve these types of problems.
Method 1:
Solve for [tex]T_1 \cdot sin(30)[/tex] and [tex]T_1 \cdot cos(30)[/tex] in both equations.
- [tex]T_1 \cdot cos(30) = T_2 \cdot cos(45)[/tex]
- [tex]T_1\cdot sin(30) = mg - T_2 \cdot sin(45)[/tex]
We know that [tex]\frac{sin\theta}{cos\theta}[/tex] = [tex]tan\theta[/tex], so we can reorder both equations and divide them.
- [tex]T_1\cdot sin(30) = mg - T_2 \cdot sin(45)[/tex]
- [tex]T_1 \cdot cos(30) = T_2 \cdot cos(45)[/tex]
- [tex]tan(30)=\frac{mg-T_2\cdot sin(45)}{T_2\cdot cos(45)}[/tex]
Note that [tex]\frac{T_1}{T_1}[/tex] cancels, leaving us with tan(30).
Now that we have this equation, we can plug in known values. We are trying to solve for m, the largest mass of the crate that the ropes can support. We know that:
- [tex]g=9.8\ \frac{m}{s^2}[/tex]
- [tex]T_2=1500\ \text{N}[/tex]
Let's substitute these values into our equation:
- [tex]tan(30)=\frac{m(9.8)-(1500)\cdot sin(45)}{(1500)\cdot cos(45)}[/tex]
Multiply 1500 * cos(45) to both sides of the equation.
- [tex]tan(30) \cdot (1500 \cdot cos(45)) =9.8m-(1500 \cdot sin(45))[/tex]
Add 1500 * sin(45) to both sides of the equation.
- [tex]tan(30) \cdot (1500 \cdot cos(45)) + 1500 \cdot sin(45)) =9.8m[/tex]
Divide both sides of the equation by 9.8.
- [tex]\frac{tan(30) \cdot (1500 \cdot cos(45)) + (1500 \cdot sin(45)) }{9.8}= m[/tex]
Simplify.
- [tex]\frac{1673.032607}{9.8}=m[/tex]
The largest mass the ropes can support is around 170 kg (round down, not up).
Method 2:
We can use the system of equations again.
- [tex]T_1 \cdot cos(30) = T_2 \cdot cos(45)[/tex]
- [tex]T_1\cdot sin(30) = mg - T_2 \cdot sin(45)[/tex]
This time, let's do a little guess and check. Using the first equation, plug in 1500 N for T2 and solve for T1.
- [tex]T_1 \cdot cos(30) = (1500) \cdot cos(45)[/tex]
Divide both sides of the equation by cos(30).
- [tex]T_1=\frac{1500 \cdot cos(45)}{cos(30)}[/tex]
Simplify.
- [tex]T_1=1224.744871 \ \text{N}[/tex]
Now plug in 1500 N for T1 and solve for T2.
- [tex](1500) \cdot cos(30) = T_2 \cdot cos(45)[/tex]
Divide both sides by cos(45).
- [tex]T_2=\frac{1500\cdot cos(30)}{cos(45)}[/tex]
- [tex]T_2=1837.117307 \ \text{N}[/tex]
Since the tension cannot exceed 1500 N without the rope breaking, we must use the first substitution, where [tex]T_1=1224.744871[/tex] and [tex]T_2=1500 \ \text{N}[/tex].
Now we can use these known values and plug them into the second equation: [tex]T_1\cdot sin(30) = mg - T_2 \cdot sin(45)[/tex]
Add [tex]T_2 \cdot sin(45)[/tex] to both sides of the equation.
- [tex]mg=T_1 \cdot sin(30) + T_2 \cdot sin(45)[/tex]
Divide both sides of the equation by g.
- [tex]m=\frac{T_1 \cdot sin(30) + T_2 \cdot sin(45)}{g}[/tex]
Plug in known values and solve for m.
- [tex]m=\frac{(1224.744871) \cdot sin(30) + (1500) \cdot sin(45)}{9.8}[/tex]
Simplify.
- [tex]m=\frac{1673.032607}{9.8}[/tex]
- [tex]m=170.717613[/tex]
As you can see, we get the same answer with either method. The largest mass the ropes can support is 170 kg.
