Car X traveled 3d distance in t time. Car Y traveled 2d distance in t time. Therefore, the speed of car X, is 3d/t, the speed of car Y, is 2d/t. Since speed is the distance taken in a given time.
In figure-2, they are at the same place, we are asked to find car Y's position when car X is at line-A. We can calculate the time car X needs to travel to there. Let's say that car X reaches line-A in t' time.
Okay, it takes t time for car X to reach line-A. Let's see how far does car Y goes.
We found that car Y travels 2d distance. So, when car X reaches line-A, car Y is just a d distance behind car X.
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:
1.65
Explanation:
The equation of the forces along the horizontal direction is:
(1)
where
F = 65 N is the force applied with the push
is the frictional force
m = 4 kg is the mass
is the acceleration
The force of friction can be written as (2), where
is the coefficient of kinetic friction
R is the normal force exerted by the floor
The equation of forces along the vertical direction is
(3)
since the bookcase is in equilibrium. Substituting (2) and (3) into (1), we find
And solving for ,