a railroad tie weights 920 N and is 2.6 m long. How much force is required to: pick it up off the ground? lift one end and rotat
1 answer:
1) The minimum force needed is 920 N
2) The minimum force is 460 N
3) The minimum force is 598 N
Explanation:
1)
We can answer this part by simply looking at the forces involved. In fact, there are two forces acting on the railroad:
- Its weight, W, acting downward
- The force applied to lift it, F, upward
So the net force on the railroad is
where
W = 920 N is the weight of the railroad
In order to lift the railroad, the net force must be upward, so
And therefore
which means that the minimum force needed is equal to the weight of the railroad, 920 N.
2)
In this case, we have to use the principle of equilibrium of moments.
In fact, when the railroad rotates uniformly (=constant angular speed) about its end, it means that the moment produced by the weight (acting in one direction) is equal to the moment produced by the force applied (acting in the other direction). Therefore, we can write:
where
W = 920 N is the weight
L = 2.6 m is the length of the railroad
F is the force applied
We wrote L/2 on the left of the equation because the weight acts at the center of mass of the railroad (located at the midpoint), while on the right it is L because the force F is applied at the end of the railroad.
Solving for F,
3)
This problem is similar to the previous part, however in this case, the force applied F is applied 0.6 m from the end, pivoting around the opposite end.
This means that the distance between the point of application of the force F and the pivot is
L' = L - 0.6
where
L = 2.6 m
Therefore the equation for the equilibrium of moments becomes
and substituting
W = 920 N
L = 2.6
We find the magnitude of F:
Learn more about forces:
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Answer:
b) a = -k / m x , c) d²x / dt² = - A w² cos (wt+Ф) , d) and e) T = 2π √m / k
h) a = - A w² cos (wt+Ф)
Explanation:
a) see free body diagram in the attachment
b) We write Newton's second law
Fe = m a
-k x = ma
a = -k / m x
c) the acceleration is
a = d²x / dt²
If x = A cos wt
v = dx / dt = -A w sin (wt +Ф)
a = d²x / dt² = - A w² cos (wt+Ф)
d) we substitute in Newton's second law
d²x / dt² = -k / m x
We call
w² = k / m
e) substitute to find w
-A w² cos (wt+Ф) = -k / m A cos (wt+Ф)
w² = k / m
Angular velocity and frequency are related
w = 2π f
f = 1 / T
We substitute
T = 2π / w
T = 2π √m / k
g) v= - A w sin (wt+Ф)
h) acceleration is
a = - A w² cos (wt+Ф)
Answer:
123.30 m
Explanation:
Given
Speed, u = 22 m/s
acceleration, a = 1.40 m/s²
time, t = 7.30 s
From equation of motion,
v = u + at
where,
v is the final velocity
u is the initial velocity
a is the acceleration
t is time
V = at + U
using equation v - u = at to get line equation for the graph of the motion of the train on the incline plane
where m is the slope
Comparing equation (1) and (2)
a = m
Since the train slows down with a constant acceleration of magnitude 1.40 m/s² when going up the incline plane. This implies the train is decelerating. Therefore, the train is experiencing negative acceleration.
a = - 1.40 m/s²
Sunstituting a = - 1.40 m/s² and u = 22 m/s
The speed of the train at 7.30 s is 11.78 m/s.
The distance traveled after 7.30 sec on the incline is the area cover on the incline under the specific interval.
Area of triangle + Area of rectangle
= 37.303 + 85.994
= 123. 297 m
≈ 123. 30 m