A 90 N boy is hanging on a rope that extends between two poles. Find the tension of the 2 ropes. The angles were 5 and 10 degrees
Your diagram probably shows more than what you stated,
e.g. are the 2 points of contact of the ropes at the same horizontal plane? I will assume they are.
Let the tension in the rope at 5° be t1
let the tension in the rope at 10° be t2
for the horizontal vectors:
t1 cos5 = t2 cos10 -----> t1 = t2 cos10/cos5 ***
for the vertical vectors:
t1 sin5 + t2 sin10 = 90
using substitution:
(t2cos10/cos5)(sin5) + t2 sin10 = 90
mutiply each term by cos5
t2 cos10 sin5 + t2 sin10 cos5 = 90cos5
t2(sin5cos10 + cos5sin10) = 90cos5
t2( sin15) = 90 cos5
t2 = 90 cos5/sin15 = ....
check my work, I did not write it out in full first.
Btw, I was using the identity :
sin(A+B) = sinAcosB + cosAsinB in the 2nd last step above to get sin 15°
once you have that , you can find t1 from ***
Sal Khan has a video of this type of problem.
It is one of his earlier ones, so it might be hard to read some of the diagrams.
https://www.khanacademy.org/science/physics/forces-newtons-laws/tension-tutorial/v/tension-part-2
THANK YOU REINY :))
To find the tension in the two ropes, we can use the concept of forces in equilibrium.
First, let's draw a diagram to visualize the problem:
```
T1
|\
| \
| \
| \
90N \
| \
| \
|_______\
|
T2
```
From the diagram, we can see that the vertical component of the force due to the tension in the ropes is equal to the weight of the boy, which is 90 N. This means that the sum of the vertical components of the tensions is equal to 90 N.
Now, let's calculate the vertical components of the tensions.
The vertical component of T1 can be calculated as:
Vertical component of T1 = T1 * sin(5°)
The vertical component of T2 can be calculated as:
Vertical component of T2 = T2 * sin(10°)
Since the sum of the vertical components of the tensions is equal to 90 N, we can write the equation:
T1 * sin(5°) + T2 * sin(10°) = 90
Now, let's solve for the tensions.
We will need one more equation to solve for the unknowns, which is the horizontal equilibrium of forces. Since there are no horizontal forces acting on the boy, the horizontal components of the tensions cancel each other out; they are in equilibrium.
Therefore, we can write:
T1 * cos(5°) = T2 * cos(10°)
Now, we have two equations:
1. T1 * sin(5°) + T2 * sin(10°) = 90
2. T1 * cos(5°) = T2 * cos(10°)
Using these equations, we can solve for the tensions.
To find the tension in the ropes, we need to break down the forces acting on the boy.
Let's consider the forces in the horizontal direction. Since there is no horizontal component of force acting on the boy, the tension in the ropes will cancel each other out.
Now, let's consider the vertical forces. There are two ropes pulling the boy upwards, while his weight is pulling him downwards. The net upward force should balance the downward force.
Using trigonometry, we can find the vertical component of the tension in each rope. We will use the angle of 5 degrees for one rope and the angle of 10 degrees for the other rope.
First, we need to find the vertical component of the weight of the boy. The weight can be found using the formula:
Weight = mass * gravity
Since the weight is given as 90 N, we can solve for the mass using the formula:
mass = Weight / gravity
Assuming the acceleration due to gravity is approximately 9.8 m/s^2, we have:
mass = 90 N / 9.8 m/s^2 = 9.18 kg (approximately)
Now, let's find the vertical component of the tension in the rope with a 5 degree angle. We'll call this tension T1.
T1 = weight / sin(angle)
T1 = (90 N) / sin(5 degrees) = 1032.66 N (approximately)
Similarly, let's find the vertical component of the tension in the rope with a 10 degree angle. We'll call this tension T2.
T2 = weight / sin(angle)
T2 = (90 N) / sin(10 degrees) = 313.27 N (approximately)
Therefore, the tension in the first rope is approximately 1032.66 N, and the tension in the second rope is approximately 313.27 N.