A uniform meter for length of 5m is suspended horizontal from two vertical p and q is attached at 2 m from the other end while q is attached at 0.8 m from one end .give that the weight of the rod is 110 n . calculate the tension in each of the string
To solve this problem, we can set up an equation using the principle of moments. The sum of the clockwise moments about any point must be equal to the sum of the anticlockwise moments.
Let's choose the point P as the pivot point. The clockwise moment due to the weight of the rod is given by the weight (110 N) multiplied by the distance from the pivot point to the center of mass of the rod, which is 2.5 m:
Clockwise moment = 110 N x 2.5 m = 275 Nm
Now let's calculate the anticlockwise moment due to the tension in the string PQ. The tension in PQ can be split into two components, one acting vertically upwards and another component acting horizontally towards the left. These components can be calculated using trigonometry.
The angle between PQ and the horizontal is given by:
θ = tan^(-1)((3/5) / (4/5)) = tan^(-1)(3/4) ≈ 36.87°
The horizontal component of the tension in PQ is:
TQ_horizontal = TQ * cos(θ)
The vertical component of the tension in PQ is:
TQ_vertical = TQ * sin(θ)
Now, the anticlockwise moment due to the horizontal component of the tension in PQ is:
Anticlockwise moment = TQ_horizontal * 0.8 m
The anticlockwise moment due to the vertical component of the tension in PQ is:
Anticlockwise moment = TQ_vertical * 4 m
To balance the moments, the sum of the anticlockwise moments must be equal to the clockwise moment:
TQ_horizontal * 0.8 m + TQ_vertical * 4 m = 275 Nm
We can rearrange this equation to solve for TQ_horizontal:
TQ_horizontal = (275 Nm - TQ_vertical * 4 m) / 0.8 m
Now let's substitute the values for TQ_horizontal and TQ_vertical:
TQ_horizontal = (275 Nm - TQ * sin(θ) * 4 m) / 0.8 m
Substituting TQ_horizontal back into the equation, we get:
(275 Nm - TQ * sin(θ) * 4 m) / 0.8 m * 0.8 m + TQ * sin(θ) * 4 m = 275 Nm
Simplifying the equation, we have:
275 Nm - TQ * sin(θ) * 4 m + TQ * sin(θ) * 4 m = 275 Nm
275 Nm = 275 Nm
As we can see, the equation is satisfied, which means that the tension in each of the strings is balanced and equal to 110 N.
To calculate the tension in each of the strings, we can use the principle of moments.
First, let's calculate the weight of the part of the rod between the pivot point and point P. Since the weight of the uniform rod is 110 N and it is suspended from the center, the weight of this part will be half of the total weight.
Weight of the rod between pivot and P = (110 N) / 2 = 55 N
Next, let's consider the moments about the pivot point. The clockwise moments from each string tension will equal the anticlockwise moment from the weight of the rod. Therefore, we can write:
(Pivot to P distance) * (Tension at P) = (Rod weight between pivot and P) * (Entire length of the meter)
Using the given values, we have:
2 * Tension at P = 55 * 5
Simplifying:
2 * Tension at P = 275
Tension at P = 275 / 2
Tension at P = 137.5 N
Now, let's calculate the tension at Q. For this, we will consider the moments about the pivot point again. The clockwise moments from the tension at point P will equal the anticlockwise moments from the weight of the rod and the tension at point Q. Therefore, we can write:
(Pivot to Q distance) * (Tension at Q) = (Rod weight between pivot and Q) * (Entire length of the meter)
Using the given values, we have:
(2 + 0.8) * Tension at Q = (110 - 55) * 5
Simplifying:
2.8 * Tension at Q = 55 * 5
2.8 * Tension at Q = 275
Tension at Q = 275 / 2.8
Tension at Q ≈ 98.21 N
Therefore, the tension at point P is approximately 137.5 N and the tension at point Q is approximately 98.21 N.