A 15 m ladder with a mass of 51 kg is leaning against a frictionless wall, which makes an angle of 60 degrees with the horizonta
l. What is the horizontal force exerted by the ground on the ladder when an 81 kg object is 4.0 m from the bottom?
2 answers:
Answer:
Explanation:
Given that, .
Mass of ladder is 51kg
Then, it weight is
WL = mg = 51 × 9.81 = 500.31N
This weight will act at the midpoint of the ladder
Length of ladder is 15m
The ladder makes an angle 55°C with the horizontal
An object whose mass is 81kg is at 4m from the bottom of the ladder
Then, weight of object
Wo = mg = 81 × 9.81 = 794.61 N
Using newton second law
Check attachment
Ng is normal force on the ground
Ff is the horizontal frictional force
Nw Is the normal force on the wall
ΣFy = 0
Ng = Wo + WL
Ng = 794.61 + 500.31
Ng = 1294.92 N
Also
ΣFx = 0
Ff — Nw = 0
Then,
Ff = Nw
Now taking moment about point A.
Check attachment
using the principle of equilibrium
Sum of clockwise moment equals to sum of anti-clockwise moment
Also note that the Normal force on the wall is not perpendicular to the ladder, so we will resolve that and also the weights of ladder and weight of object
Clockwise = Anticlockwise
Wo•Cos60 × 4 + WL•Cos60 × 7.5 = Nw•Sin60 × 15
794.61Cos60 × 4 + 500.31Cos60 × 7.5 = Nw × Sin60 × 15
1589.22 + 1876.163 = 12.99•Nw
3465.383 = 12.99•Nw
Nw = 3465.383 / 12.99
Nw = 266.77 N
Since, Nw = Ff
Then, Ff = 266.77N
the horizontal force exerted by the ground on the ladder is 266.77 N
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Answer:
(a) W = 650J
(b) Wf = 529.2J
(c) W = 0J
(d) W = 0J
(e) ΔK = 120.8J
(f) v2 = 2.58 m/s
Explanation:
(a) In order to find the work done by the applied force you use the following formula:
(1)
F: applied force = 130N
d: distance = 5.0m
The work done by the applied force is 650J
(b) The increase in the internal energy of the box-floor system is given by the work done of the friction force, which is calculated as follow:
(2)
μ: coefficient of friction = 0.300
M: mass of the box = 36.0kg
g: gravitational constant = 9.8 m/s^2
The increase in the internal energy is 529.2J
(c) The normal force does not make work on the box because the normal force is perpendicular to the motion of the box.
(d) The same for the work done by the normal force. The work done by the gravitational force is zero because the motion of the box is perpendicular o the direction of the gravitational force.
(e) The change in the kinetic energy is given by the net work on the box. You use the following formula:
(3)
You calculate the total work:
(4)
F: applied force = 130N
Ff: friction force
d: distance = 5.00m
The friction force is:
Next, you replace the values of all parameters in the equation (4):
The change in the kinetic energy of the box is 120.8J
(e) The final speed of the box is calculated by using the equation (3):
(5)
v2: final speed of the box
v1: initial speed of the box = 0 m/s
You solve the equation (5) for v2:
The final speed of the box is 2.58m/s