A plank with m=0.12 kg and 50cm long has one mass of m1=0.11kg sitting on one end and a m2=0.055kg at the other end. Where should the pivot point be to balance the plank?
I figured the equation out
Xcg=(m1x1+m2x2+m3(L/2))/(m1+m2+m3)
m1 = 0.055kg
x1 = 0cm
m2 = 0.11kg
x2 = 50cm
for the plank you need to add
m3 = 0.12kg
and distance is L=50cm
it calculates to Xcg=29.825cm
To determine the pivot point where the plank will be balanced, we can use the principle of moments. The principle of moments states that for an object to be in rotational equilibrium, the sum of the clockwise moments must equal the sum of the anticlockwise moments.
In this case, let's assume that the pivot point is at a distance x from the end where the m1 is placed.
The weight of the plank itself can be considered to act at its center of mass. Since the mass of the plank is m = 0.12 kg and its length is 50 cm, the center of mass is located at the midpoint, which is 25 cm from both ends.
The weight of the plank can be calculated as follows:
Weight of plank (W_plank) = mass of plank (m) x acceleration due to gravity (g)
W_plank = 0.12 kg x 9.8 m/s^2
W_plank = 1.176 N
Now, let's calculate the clockwise and anticlockwise moments:
Clockwise moments:
The moment due to m1 = m1 x g x (x - 50 cm)
The moment due to m2 = m2 x g x (x)
Anticlockwise moments:
The moment due to the weight of the plank = W_plank x (25 cm)
According to the principle of moments, the sum of the clockwise moments must equal the sum of the anticlockwise moments:
m1 x g x (x - 50 cm) + m2 x g x (x) = W_plank x (25 cm)
Substituting the known values:
0.11 kg x 9.8 m/s^2 x (x - 50/100 m) + 0.055 kg x 9.8 m/s^2 x (x) = 1.176 N x (25/100 m)
Simplifying the equation:
(0.1078 x (x - 0.5)) + (0.0539 x x) = 0.294
Expanding and rearranging:
0.1078x - 0.0539 + 0.0539x = 0.294
0.1617x = 0.3479
x ≈ 2.151 m
Therefore, the pivot point should be located approximately 2.151 meters from the end where the m1 mass is placed to balance the plank.