A uniform wooden metre rule of mass 90g is pivoted at the 40cm mark.if the rule is in equilibrium with an unknown mass M placed at the 10cm mark and 72g mass at the 70cm mark ,determine M.(WAEC 2000)
This is not possible to solve. Unless we know how long the rod is, we have no idea how far from the fulcrum the mass of the rod is distributed. Just for simplicity, let's assume that the pivot is at the center, so the rod is 80cm long.
Now consider the rod as massless, and having two extra masses of 40g at the 20cm mark, and 50g at the 60cm mark. For the forces to balance, we need
30M + 20*40 = 20*50 + 30*72
Now just solve for M.
oops. I was dividing the mass 40:50 instead of evenly. Try
30M + 20*45 = 20*45 + 30*72
To determine the unknown mass M, we can use the principle of moments. The principle of moments states that for a system in equilibrium, the sum of the clockwise moments is equal to the sum of the anti-clockwise moments.
In this case, let's consider the pivot point at the 40cm mark as the reference point. The clockwise moments will be negative, and the anti-clockwise moments will be positive.
First, let's calculate the moments caused by the 72g mass at the 70cm mark:
Moment of the 72g mass = Mass x Distance from the pivot
= 72g x (70cm - 40cm)
= 72g x 30cm
= 2160 g·cm (clockwise)
Next, consider the unknown mass M at the 10cm mark:
Moment of the unknown mass M = Mass x Distance from the pivot
= M x (10cm - 40cm)
= M x (-30cm)
= -30M g·cm (clockwise)
Since the system is in equilibrium, the sum of the moments is zero:
Sum of clockwise moments = Sum of anti-clockwise moments
-30M g·cm + 2160 g·cm = 0
Simplifying the equation:
-30M + 2160 = 0
30M = 2160
M = 2160 / 30
By dividing 2160 by 30, we find that M is equal to 72 grams.
Therefore, the unknown mass M is 72 grams.