2. A bar of uniform section is 81.4cm long and weighs 4.45 kg. A weight of 12.8 kg is suspended from one end. The bar weight combination is to be suspended from a cable attached at the balance point.
a)Determine how far from the weight should the cable be attached.
b)Calculate the tension, in newtons N, in the cable.
To determine how far from the weight the cable should be attached, you need to find the balance point of the bar. The balance point is the point on the bar where the torque on one side of the bar is equal to the torque on the other side.
The torque is given by the product of the weight and the distance from the point of suspension. In this case, the torque on the left side of the bar is the weight of the bar (4.45 kg) multiplied by the distance x, while the torque on the right side of the bar is the weight attached to the end of the bar (12.8 kg) multiplied by the length of the bar (81.4 cm - x).
So, the equation for the balance of torques is:
4.45 kg * x = 12.8 kg * (81.4 cm - x)
Now, let's solve for x:
4.45 kg * x = 12.8 kg * 81.4 cm - 12.8 kg * x
4.45 kg * x + 12.8 kg * x = 12.8 kg * 81.4 cm
(4.45 kg + 12.8 kg) * x = 12.8 kg * 81.4 cm
17.25 kg * x = 1041.92 kg*cm
Now, divide both sides by 17.25 kg:
x = 1041.92 kg*cm / 17.25 kg
x ≈ 60.48 cm
So, the cable should be attached approximately 60.48 cm from the weight.
To calculate the tension in the cable, we need to consider the forces acting on the bar. Since the bar is in equilibrium, the sum of the forces in the vertical direction is zero.
The weight of the bar itself (4.45 kg) and the weight attached to the end (12.8 kg) both act downward. The tension in the cable acts upward. So, we have two downward forces and one upward force.
The tension in the cable can be calculated as the sum of the two downward forces:
Tension = Weight of the bar + Weight attached to the end
Tension = (4.45 kg + 12.8 kg) * 9.8 m/s^2 (acceleration due to gravity)
Tension ≈ 177.67 N
Therefore, the tension in the cable is approximately 177.67 N.