Auniform bar weighing 40newton and measuring 2metres is hinged at one end so that the unhinged end can moveupanddown.Ifitisheldhorizontallybyaspringbalanceat0.3metresfromthe free end,determine the reading of the spring balance.
To determine the reading of the spring balance, we need to consider the balance of forces acting on the uniform bar when it is held horizontally.
First, let's analyze the forces acting on the bar:
1. Weight of the bar: The bar weighs 40 Newtons. This force acts downwards and can be considered as acting at the center of gravity of the bar. Since the bar is uniform, the center of gravity is at its midpoint, which is 1 meter from the hinge.
2. Reaction force at the hinge: The hinge causes a reaction force that acts upwards to balance the weight of the bar.
3. Force exerted by the spring balance: The spring balance exerts a force upwards at a distance of 0.3 meters from the free end of the bar.
To find the reading of the spring balance, we need to find the balancing force exerted by the spring balance.
Considering the rotational equilibrium of the bar about the hinge point, we can write:
Clockwise moment = Anti-clockwise moment
The clockwise moment is the torque created by the weight of the bar, which can be calculated as:
Torque = Weight × Distance from the hinge = 40 N × 1 m = 40 N·m
The anti-clockwise moment is the torque created by the force exerted by the spring balance, which can be calculated as:
Torque = Force × Distance from the hinge = Reading of the spring balance × 0.3 m
Since the bar is held horizontally, these two torques must be equal:
40 N·m = Reading of the spring balance × 0.3 m
To find the reading of the spring balance, we can rearrange the equation:
Reading of the spring balance = (40 N·m) / (0.3 m)
Reading of the spring balance ≈ 133.33 N
Therefore, the reading of the spring balance is approximately 133.33 Newtons.