To solve this problem we will apply the concepts related to equilibrium, for this specific case, through the sum of torques.

If the distance in which the 600lb are applied is 6in, we will have to add the unknown Force sum, at a distance of 27in - 6in will be equivalent to that required to move the object. So,



So, Force that must be applied at the long end in order to lift a 600lb object to the short end is 171.42lb
Answer:
0.087 m
Explanation:
Length of the rod, L = 1.5 m
Let the mass of the rod is m and d is the distance between the pivot point and the centre of mass.
time period, T = 3 s
the formula for the time period of the pendulum is given by
.... (1)
where, I is the moment of inertia of the rod about the pivot point and g is the acceleration due to gravity.
Moment of inertia of the rod about the centre of mass, Ic = mL²/12
By using the parallel axis theorem, the moment of inertia of the rod about the pivot is
I = Ic + md²

Substituting the values in equation (1)


12d² -26.84 d + 2.25 = 0


d = 2.15 m , 0.087 m
d cannot be more than L/2, so the value of d is 0.087 m.
Thus, the distance between the pivot and the centre of mass of the rod is 0.087 m.
Seven days after a waxing gibbous moon, the moon will be waning gibbous, and at some point during that seven days, it will have been Full.
Answer:
The constriction causes the mercury column to break under tension, leaving a vacuum between the bottom of the column and that in the bulb, and the top of the column stays still at the position reached in the body - a "peak hold" system.