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

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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.