1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
DerKrebs [107]
3 years ago
7

A 54 kg person stands on a uniform 20 kg, 4.1 m long ladder resting against a frictionless wall.

Physics
1 answer:
SVETLANKA909090 [29]3 years ago
7 0

A) Force of the wall on the ladder: 186.3 N

B) Normal force of the ground on the ladder: 725.2 N

C) Minimum value of the coefficient of friction: 0.257

D) Minimum absolute value of the coefficient of friction: 0.332

Explanation:

a)

The free-body diagram of the problem is in attachment (please rotate the picture 90 degrees clockwise). We have the following forces:

W=mg: weight of the ladder, with m = 20 kg (mass) and g=9.8 m/s^2 (acceleration of gravity)

W_M=Mg: weight of the person, with M = 54 kg (mass)

N_1: normal reaction exerted by the wall on the ladder

N_2: normal reaction exerted by the floor on the ladder

F_f = \mu N_2: force of friction between the floor and the ladder, with \mu (coefficient of friction)

Also we have:

L = 4.1 m (length of the ladder)

d = 3.0 m (distance of the man from point A)

Taking the equilibrium of moments about point A:

W\frac{L}{2}sin 21^{\circ}+W_M dsin 21^{\circ} = N_1 Lsin 69^{\circ}

where

Wsin 21^{\circ} is the component of the weight of the ladder perpendicular to the ladder

W_M sin 21^{\circ} is the component of the weight of the man perpendicular to the ladder

N_1 sin 69^{\circ} is the component of the normal  force perpendicular to the ladder

And solving for N_1, we find the force exerted by the wall on the ladder:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{mg}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+Mg\frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{(20)(9.8)}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+(54)(9.8)\frac{3.0}{4.1}\frac{sin 21^{\circ}}{sin 69^{\circ}}=186.3 N

B)

Here we want to find the magnitude of the normal force of the ground on the ladder, therefore the magnitude of N_2.

We can do it by writing the equation of equilibrium of the forces along the vertical direction: in fact, since the ladder is in equilibrium the sum of all the forces acting in the vertical direction must be zero.

Therefore, we have:

\sum F_y = 0\\N_2 - W - W_M =0

And substituting and solving for N2, we find:

N_2 = W+W_M = mg+Mg=(20)(9.8)+(54)(9.8)=725.2 N

C)

Here we have to find the minimum value of the coefficient of friction so that the ladder does not slip.

The ladder does not slip if there is equilibrium in the horizontal direction also: that means, if the sum of the forces acting in the horizontal direction is zero.

Therefore, we can write:

\sum F_x = 0\\F_f - N_1 = 0

And re-writing the equation,

\mu N_2 -N_1 = 0\\\mu = \frac{N_1}{N_2}=\frac{186.3}{725.2}=0.257

So, the minimum value of the coefficient of friction is 0.257.

D)

Here we want to find the minimum coefficient of friction so the ladder does not slip for any location of the person on the ladder.

From part C), we saw that the coefficient of friction can be written as

\mu = \frac{N_1}{N_2}

This ratio is maximum when N1 is maximum. From part A), we see that the expression for N1 was

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{d}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}

We see that this quantity is maximum when d is maximum, so when

d = L

Which corresponds to the case in which the man stands at point B, causing the maximum torque about point A. In this case, the value of N1 is:

N_1 = \frac{W}{2}\frac{sin 21^{\circ}}{sin 69^{\circ}}+W_M \frac{L}{L}\frac{sin 21^{\circ}}{sin 69^{\circ}}=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{W}{2}+W_M)

And substituting, we get

N_1=\frac{sin 21^{\circ}}{sin 69^{\circ}}(\frac{(20)(9.8)}{2}+(54)(9.8))=240.8 N

And therefore, the minimum coefficient of friction in order for the ladder not to slip is

\mu=\frac{N_1}{N_2}=\frac{240.8}{725.2}=0.332

Learn more about torques and equilibrium:

brainly.com/question/5352966

#LearnwithBrainly

You might be interested in
Oil having a density of 922kg/m^3 floats on water. A rectangular block of wood 3.97 cm high and with a density of 963 kg/m^3 flo
Blizzard [7]

Explanation:

For the equilibrium:

\rho_{wood}gh-\rho_{oil}g(h-x)-\rho_{water}gx=0ρ

wood

gh−ρ

oil

g(h−x)−ρ

water

gx=0

\rho_{wood}h-\rho_{oil}(h-x)-\rho_{water}x=0ρ

wood

h−ρ

oil

(h−x)−ρ

water

x=0

(974)(3.97)-928(3.97-x)-1000x=0(974)(3.97)−928(3.97−x)−1000x=0

x=2.54\ cmx=2.54 cm

3 0
2 years ago
A small airplane has to reach a speed if 27.8 m/s to takeoff. It can accelerate at 2.00 m/s^2. What is the minimal length of run
pickupchik [31]
Using the constant acceleration formula v^2 = u^2 + 2as, we can figure out that it would take a distance of 193.21m to reach 27.8m/s

3 0
3 years ago
Many of water's emergent properties, such as its cohesion, its high specific heat, and its high heat of vaporization, result fro
Bingel [31]

Answer:

Option 5.

Explanation:

Many of the properties of water like high specific heat, cohesion, high vaporization heat, etc can be contributed to the polar nature of water molecule.

Water being a polar molecule as it contains positively charged hydrogen and an electro-negative oxygen which results in uneven or non uniformity in sharing of electrons which leads to dipole formation and hence polarization of the molecule due to which it attracts its neighboring molecules.

This polar nature imparts the properties like cohesion, surface tension , adhesion, etc due to the presence of hydrogen bonds in water molecule.

3 0
3 years ago
Object A attracts object B with a gravitational force of 5 newtons from a given distance. If the distance between the two object
enyata [817]

Answer:

10 newtons, because the gravitaional force willbe stronger the closer it gets.

7 0
3 years ago
one of the ways, covered in the article and in class, that we charged an object was rubbing it against carpet on the floor true
astraxan [27]

Answer:

It is possible to statically charge objects by rubbing it against carpet fibers, but I'm not sure if that was in the article that you read.

Explanation:

Static charge can build up via carpet fibers.

5 0
2 years ago
Other questions:
  • Calculate the force at sea level that a boy of mass 50 kg exerts on a chair in which he is sitting
    9·1 answer
  • A train moving west with an initial velocity of 20 m/s accelerates at 4 m/s2 for 10 seconds. During this time, the train moves a
    13·2 answers
  • Given that the distance from the left end of the string to the first antinode is 27.5 cm , calculate the wavelength of the stand
    7·1 answer
  • Where must an object be placed to form an image 30.0 cm from a diverging lens with a focal length of 43.0 cm?
    6·1 answer
  • You stand at the top of a deep well. To determine the depth, D, of the well you drop a rock from the top of the well and listen
    11·1 answer
  • Why is gravitational forces equal for everyone?
    12·1 answer
  • Hi there hope your having a great day!! my questions both are SCIENCE laws of motion related fyi
    5·1 answer
  • As the air on the surface of the earth warms what happens to the density of the air
    7·1 answer
  • A child and sled with a combined mass of 54.8
    11·1 answer
  • State Newton's second law of motion in terms of momentum<br>​
    7·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!