1. A Uniform rod AB 3m long weighing 3 Kg is held horizontally by two strong vertical strings at its ends. A weight of 7 kg is hanging on the rod at a distance of 1m from end A and another weight of 12 Kg is hanging at 0.5 m from end B.(a)Make the free body diagram of the rod and (b)Find the tension in two strings.
you know the tension sum is equal to the weights.
Ta+Tb-3*g-12*g-7g=0 where g is 9.8N/kg
Then, sum moments about any point. I will sum at end a.
1.5*3g -Tb*3+1*7*g+2.5*12g=0
(a) To make the free body diagram of the rod, we need to consider the forces acting on the rod.
At end A, there is a tension force from string A pulling upwards. At end B, there is a tension force from string B pulling upwards. Additionally, there are two downward forces acting on the rod – the weight hanging at a distance of 1m from end A, and the weight hanging at a distance of 0.5m from end B.
So, the free body diagram of the rod will look like this:
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| |
| |
| Tension A |
| |
| W1 |
| |
| |
|-------------------|
| |
| W2 |
| |
| Tension B |
| |
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(b) Now, we need to find the tension in the two strings.
To find the tension in string A, we need to consider the forces acting on the rod in the vertical direction.
Let's assume the tension in string A is T_a and the tension in string B is T_b.
At end A, the vertical forces are: T_a (upwards) and the weight W1 (downwards).
So, we can write the equation in the vertical direction:
T_a - W1 = 0
At end B, the vertical forces are: T_b (upwards) and the weight W2 (downwards).
So, we can write the equation in the vertical direction:
T_b - W2 = 0
Now, we can substitute the values for W1, W2, and solve the equations to find the tensions T_a and T_b.
Given:
Length of the rod, AB = 3m
Weight of the rod = 3Kg
Weight hanging at a distance of 1m from end A = 7kg
Weight hanging at a distance of 0.5m from end B = 12kg
Using the equation for the center of mass of a uniform rod,
Center of mass (x) = (m1 * x1 + m2 * x2) / (m1 + m2)
where m1 and m2 are the weights and x1 and x2 are the distances from the center of mass.
We can calculate the center of mass, x:
x = (7 * 1 + 12 * 0.5) / (7 + 12) = 0.84 meters
Knowing the center of mass, we can calculate the distance between the center of mass and the ends of the rod:
Distance from center of mass to end A = 0.84 m
Distance from center of mass to end B = 3 - 0.84 = 2.16 m
Now, we can calculate the moments about the two ends of the rod:
For end A, taking moments about A:
T_a * 0 + 0.84 * 7g + 2.16 * 12g = 0
where g is the acceleration due to gravity (9.8 m/s^2)
For end B, taking moments about B:
T_b * 3 + 2.16 * 12g + 0.84 * 7g = 0
Simplifying the equations, we can solve for T_a and T_b.