Question

A 300 g walking stick is placed on a pivot and balanced by a 100 g mass, as shown in

Figure 5.20. Calculate the distance of the centre of gravity of the stick from the pivot.​

Answer 1

Purplemath

The first term in the binomial is "x2", the second term in "3", and the power n is 6, so, counting ... x12 + 18x10 + 135x8 + 540x6 + 1215x4 + 1458x2 + 729 ... So memorize the Theorem and get the easy points.

Answer 2

the distance of the center of gravity of the stick from

Δ = 5 cm to make it equilibrium or balance.

Step-by-step explanation:

100 g

______Π___________________________________)

|<--- 15-----Δ--- x--->|

weight of stick (w) = 300 g acting on center with x distance from support Δ.

to get the x distance to make stick and a block balance, take moment from support.

ΣM at Δ = 0

0 = 100 (15) - 300(x)

0 = 1500 - 300x

300x = 1500

x = 1500 / 300

x = 5

therefore, the distance of the center of gravity of the stick from Δ = 5 cm

to make it equilibrium or balance.