In the figure below PT is a uniform meter rule pivoted at r the 70cm mark. Two forces 0.1N and 0.4N are applied at Q the 60cm mark and S the 85cm mark. If the meter rule is kept in equilibrium by the forces and its weight calculate the weight of the meter rule
To calculate the weight of the meter rule, we need to consider the equilibrium of forces acting on it. The forces acting on the meter rule are the two applied forces at points Q and S and the weight of the meter rule itself.
In equilibrium, the sum of the clockwise moments is equal to the sum of the anticlockwise moments. We can calculate the moments using the equation:
Moment = Force × Distance
First, let's calculate the moments of the applied forces. The force at Q is 0.1N, and it acts at 60cm. So the moment of the force at Q is:
Moment(Q) = 0.1N × 60cm
Similarly, the force at S is 0.4N, and it acts at 85cm. So the moment of the force at S is:
Moment(S) = 0.4N × 85cm
Next, we need to consider the moment due to the weight of the meter rule. The weight of an object is given by the equation:
Weight = Mass × Gravitational Acceleration
Since the meter rule is uniform, we can assume its weight acts at the center of mass, which is at the 50cm mark. The distance of the weight from the pivot point is 70cm - 50cm = 20cm.
Now, we need to find the mass of the meter rule. Since it is uniform, we can assume the mass is evenly distributed throughout its length. The meter rule has a length of 100cm and a weight of W. So the mass of 1cm of the meter rule is:
Mass of 1cm = W / 100g
The gravitational acceleration is approximately 9.8 m/s^2.
Now, we can calculate the moment due to the weight:
Moment(weight) = Weight × Distance
= (Mass of 1cm × g) × Distance of weight from pivot
= ((W / 100g) × 9.8 m/s^2) × 20cm
In equilibrium, the sum of the clockwise and anticlockwise moments is zero:
Moment(Q) + Moment(S) + Moment(weight) = 0
Now we can substitute the known values into the equation and solve for the weight of the meter rule (W).