a string ABCD is attached to two fixed points A and D, has two equal weights of 1000N attached to it at B and C. the weight rest with portions AB and CD inclined at angles of 30 degrees and 60 degrees respectively to the vertical. find the tension in portions AB, BC and CD if the inclination of the portion BC with the vertical is 120 degrees

Need answer plz

Oh, boy! Sounds like you've got a string with some serious tension going on! Let's break it down, shall we?

First, let's tackle the tension in portion AB. Since the angle of inclination is 30 degrees, we can utilize some trigonometry magic. The weight of 1000N is acting vertically downwards, so we can think of it as the hypotenuse of a right triangle. The opposite side (AB) is the tension we're looking for. Using the sine function, we have:

sin(30 degrees) = AB / 1000N

Solving for AB, we get:

AB = sin(30 degrees) * 1000N

Now, let's move on to portion BC. The angle of inclination here is 120 degrees, which means we need to switch things up a bit. We'll use the cosine function to find the tension in BC:

cos(120 degrees) = BC / 1000N

Simplifying, we get:

BC = cos(120 degrees) * 1000N

Lastly, let's get the tension in portion CD. The angle of inclination is 60 degrees, so we'll use the sine function again:

sin(60 degrees) = CD / 1000N

Solving for CD, we find:

CD = sin(60 degrees) * 1000N

And there you have it! The tension in portion AB is sin(30 degrees) * 1000N, the tension in portion BC is cos(120 degrees) * 1000N, and the tension in portion CD is sin(60 degrees) * 1000N. Time to get those tensions all dressed up and ready to go!

To find the tension in portions AB, BC, and CD, we can apply the principles of equilibrium.

Let's label the tension in AB as T_AB, the tension in BC as T_BC, and the tension in CD as T_CD.

First, let's consider the forces acting on point B:
- The weight of 1000N acting vertically downwards.
- The tension T_AB acting upwards at an angle of 30 degrees to the vertical.
- The tension T_BC acting horizontally towards point C.

Since the point B is in equilibrium, the sum of the forces in the vertical direction should be zero:
T_AB * sin(30 degrees) = 1000N

Now, let's consider the forces acting on point C:
- The tension T_BC acting horizontally towards point B.
- The weight of 1000N acting vertically downwards.
- The tension T_CD acting downwards at an angle of 60 degrees to the vertical.

Again, since point C is in equilibrium, the sum of the forces in the horizontal direction should be zero:
T_BC * cos(120 degrees) = T_CD * cos(60 degrees)

Finally, let's consider the forces acting on point D:
- The weight of 1000N acting vertically downwards.
- The tension T_CD acting downwards at an angle of 60 degrees to the vertical.

In this case, the sum of the forces in the vertical direction should be zero:
T_CD * sin(60 degrees) = 1000N

Now, we can solve these equations simultaneously to find the values of T_AB, T_BC, and T_CD.

From the first equation: T_AB = 1000N / sin(30 degrees) = 2000N

From the second equation: T_BC * cos(120 degrees) = T_CD * cos(60 degrees)
T_BC * (-0.5) = T_CD * 0.5
T_BC = -T_CD

Substituting the value of T_BC into the third equation: T_CD * sin(60 degrees) = 1000N
T_CD = 1000N / sin(60 degrees) = 1154.7N

Therefore, T_AB = 2000N, T_BC = -1154.7N, and T_CD = 1154.7N.
Note that the negative sign in T_BC indicates the opposite direction to the positive axis.

To find the tension in the different portions of the string, we can use the principles of equilibrium.

Let's start by analyzing the forces acting on the system.

1. Tension in portion AB (T_AB):
- This portion is inclined at an angle of 30 degrees to the vertical.
- There are two equal weights of 1000N each attached at B and C.
- The total weight of 2000N acts vertically downward at the center of BC.
- As there is no net force in the horizontal direction, the tension in AB should balance the horizontal component of the weight.
- The horizontal component of the weight is given by W_horizontal = W * sin(30), where W is the total weight.

Therefore, T_AB = W_horizontal = 2000N * sin(30) = 1000N.

2. Tension in portion BC (T_BC):
- This portion makes an angle of 120 degrees with the vertical.
- The weight of 2000N acts vertically downward at the center of BC.
- As there is no net force in the horizontal direction, the tension in BC should balance the horizontal component of the weight.
- The horizontal component of the weight is given by W_horizontal = W * sin(120), where W is the total weight.

Therefore, T_BC = W_horizontal = 2000N * sin(120) = 2000N * (√3/2) = 1000√3N.

3. Tension in portion CD (T_CD):
- This portion is inclined at an angle of 60 degrees to the vertical.
- There are two equal weights of 1000N each attached at B and C.
- The total weight of 2000N acts vertically downward at the center of BC.
- As there is no net force in the horizontal direction, the tension in CD should balance the horizontal component of the weight.
- The horizontal component of the weight is given by W_horizontal = W * sin(60), where W is the total weight.

Therefore, T_CD = W_horizontal = 2000N * sin(60) = 2000N * (√3/2) = 1000√3N.

So, the tension in portion AB is 1000N, the tension in portion BC is 1000√3N, and the tension in portion CD is also 1000√3N.