A UNIFORM METER RULE IS PIVOTED AT P IN 80cm mark on the meter rule force 0.2 And 0.8N Are Applied At M The 60Cm Mark And Q The 90Cm Mark.If The Rule Is In Equilibrium .Find The Mass Of The Rule.Take g=10ms-2
m=w/g, w=0.2*0.8=0.16. m=0.16/10=0.016g
Please solve the problem
To find the mass of the rule, we need to first understand the concept of equilibrium. In equilibrium, the total clockwise moment is equal to the total anticlockwise moment.
Let's break down the given information:
- The meter rule is pivoted at point P.
- The force of 0.2N is applied at point M (60cm mark).
- The force of 0.8N is applied at point Q (90cm mark).
- The rule is in equilibrium.
Now, let's calculate the moments.
The moment (M) of a force about a point is given by the formula: M = F * d
- The moment about point P due to the force at point M is: M1 = 0.2N * 80cm = 16 Ncm
- The moment about point P due to the force at point Q is: M2 = 0.8N * 10cm = 80 Ncm
Since the rule is in equilibrium, the total clockwise moment (M1) is equal to the total anticlockwise moment (M2).
Therefore, 16 Ncm = 80 Ncm
To find the mass of the rule, we can use the formula for gravitational force:
F = m * g
Where:
- F is the force (weight),
- m is the mass,
- g is the acceleration due to gravity (given as 10 ms^-2).
Since we know the force is equal to the weight, we can write:
0.2N = m * 10 ms^-2
Rearranging the equation to solve for mass (m):
m = 0.2N / 10 ms^-2
m = 0.02 kg
Therefore, the mass of the rule is 0.02 kg.