PT IS A UNIFORM METRE RULE PIVOTED ARE THE 70CM MARK THE FORCES 0.1N AND 0.4N ARE A Q THE 60CM AND S THE 87CM MARK. IF THE METRE RULE IS KEPT IN EQUILIBRIUM BY THE FORCE AND ITS WEIGHT THEN THE WEIGHT OF THE METRE RULE
To determine the weight of the meter rule, we need to first understand the concept of equilibrium and the conditions that need to be satisfied for it to occur.
In this scenario, the meter rule is in equilibrium, which means that the sum of all the forces acting on it must be zero. There are two forces acting on the rule: the applied force (Q) and the weight of the meter rule itself.
Let's break down the problem step by step:
1. Identify the forces:
- The force (F) acting at the 70 cm mark is 0.1 N.
- The force (S) acting at the 87 cm mark is 0.4 N.
2. Assign signs to the forces:
- Since the force at Q is pushing upwards, it will be considered positive.
- Since the force at S is pushing downwards, it will be considered negative.
- The weight of the meter rule will also act downwards, so it will also be considered negative.
3. Apply the condition for equilibrium:
According to the principle of equilibrium, the sum of all the forces acting on an object in equilibrium should be zero. Therefore, we can write the equation as follows:
F + S + Weight = 0
4. Substitute the known values:
- F = 0.1 N (positive force at Q)
- S = -0.4 N (negative force at S)
Substituting these into the equation, we get:
0.1 N + (-0.4 N) + Weight = 0
5. Solve for the weight of the meter rule:
Rearranging the equation, we have:
Weight = -0.1 N + 0.4 N
Weight = 0.3 N
Therefore, the weight of the meter rule is 0.3 N.