Answer:
F = 24 N
Explanation:
In this exercise we have a bar l = 100 m with a center of gravity x = 4 m, which force is needed to lift it from the other end
Let's use the rotational equilibrium relationship, where we consider the counterclockwise rotations as positive and fix the reference system at the point closest to the center of gravity
∑ τ = 0
F l -x W = 0
F =
let's calculate
F = 4/100 600
F = 24 N
Answer:
a)
a = 2 [m/s^2]
b)
a = 1.6 [m/s^2]
c)
xt = 2100 [m]
Explanation:
In order to solve this problem we must use kinematics equations. But first we must identify what kind of movement is being studied.
a)
When the car moves from rest to 40 [m/s] by 20 [s], it has a uniformly accelerated movement, in this way we can calculate the acceleration by means of the following equation:
where:
Vf = final velocity = 40 [m/s]
Vi = initial velocity = 0 (starting from rest)
a = acceleration [m/s^2]
t = time = 20 [s]
40 = 0 + (a*20)
a = 2 [m/s^2]
The distance can be calculates as follows:
where:
x1 = distance [m]
40^2 = 0 + (2*2*x1)
x1 = 400 [m]
Now the car maintains its speed of 40 [m/s] for 30 seconds, we must calculate the distance x2 by means of the following equation, it is important to emphasize that this movement is at a constant speed.
v = x2/t2
where:
x2 = distance [m]
t2 = 30 [s]
x2 = 40*30
x2 = 1200 [m]
b)
Immediately after a change of speed occurs, such that the previous final speed becomes the initial speed, the new Final speed corresponds to zero, since the car stops completely.
Note: the negative sign of the equation means that the car is stopping, i.e. slowing down.
0 = 40 - (a *25)
a = 40/25
a = 1.6 [m/s^2]
The distance can be calculates as follows:
0 = (40^2) - (2*1.6*x3)
x3 = 500 [m]
c)
Now we sum all the distances calculated:
xt = x1 + x2 + x3
xt = 400 + 1200 + 500
xt = 2100 [m]