Ask question

Ask question

All categories

2 years ago

8

An 8.0 m, 240 N uniform ladder rests against a smooth wall. The coefficient of static friction between the ladder and the ground

is 0.55, and the ladder makes a 50.0° angle with the ground. How far up the ladder can an 710 N person climb before the ladder begins to slip?

1 answer:

sweet [91]2 years ago

7 0

Answer:

5.7 m

Explanation:

AD = length of the ladder = L = 8 m

AB = distance of the center of mass of the ladder = (0.5) L = (0.5) 8 = 4 m

AC = distance of person on the ladder from the bottom end = x

W = weight of the ladder = 240 N

$F_{g}$ = weight of the person = 710 N

F = force by the wall on the ladder

N = normal force by ground on the ladder = ?

Using equilibrium of force along the vertical direction

N = $F_{g}$ + W

N = 710 + 240

N = 950 N

μ = Coefficient of static friction = 0.55

f =static frictional force on the ladder

Static frictional force is given as

f = μ N

f = (0.55) (950)

f = 522.5 N

Force equation along the horizontal direction is given as

F = f

F = 522.5 N

using equilibrium of torque about point A

F Sin50 (AD) = W Cos50 (AB) + ( $F_{g}$ Cos50 (AC))

(522.5) Sin50 (8) = (240) Cos50 (4) + (710) Cos50 (x)

x = 5.7 m

You might be interested in

A 1.5 m cylinder of radius 1.1 cm is made of a complicated mixture materials. Its resistivity depends on the distance x from the

Elis [28]

Answer:

a) $R = 171$ μΩ

b) $E = 1.7 *10^{-4} V/m$

c) $R_{2} = 1.16 *10^{-4}$ Ω

here * stand for multiplication

Explanation:

length of cylinder = 1.5 m

radius of cylinder = 1.1 cm

resistivity depends on the distance x from the left

$p(x)=a+bx^2$ ............(i)

using equation

$R = \frac{pl}{a}$

let dR is the resistance of thickness dx

$dR =\frac{p(x)dx}{a}$

where p(x) is resistivity l is length

a is area

$\int\limits^R_0 {dR} =\frac{1}{\pi r^2} \int\limits^L_0 {(a+bx^2)} \, dx \\$ .........................(2)

after integration

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$ ...............(3)

it is given $p(0) = a = 2.25 * 10 ^{-8}$ Ωm

$p(L) = a + b(L)^2$ = $8.5 * 10 ^{-8}$ Ωm

$8.5 * 10 ^{-8} = 2.25 * 10^{-8}+b(1.5)^2\\$

(here * stand for multiplication )

on solving we get

$b = 2.78* 10^{-8}$ Ωm

put each value of a and b and r value in equation 3rd we get

$R = \frac{[aL+\frac{bL^3}{3}] }{\pi r^2}$

$R = 1.71 * 10^{-4}$ Ω

$R = 171$ μΩ

FOR (b)

for mid point x = L/2

E = p(x)L

for x = L/2

$p(L/2) = a+b(L/2)^2$

for given current I = 1.75 A

so electric field

$E = \frac{[a+b(L/2)^2]I }{\pi r^2}$

by substitute the values

we get;

$E = 1.7 *10^{-4} V/m$

(here * stand for multiplication )

c ).

75 cm means length will be half

that is x = L/2

integrate the second equation with upper limit L/2

Let resistance is $R_{1}$

so after integration we get

$R_{1} = \frac{[a(L/2) +(b/3)(L^3/8)]}{\pi r^2}$

substitute the value of a , b and L we get

$R_{1} = 5.47 * 10 ^{-5}$ Ω

for second half resistance

$R_{2} = R- R_{1}$

$R_{2} = 1.7 *10^{-4} -5.47 *10^{-5}$

$R_{2} = 1.16 *10^{-4}$ Ω

(here * stand for multiplication )

5 0

2 years ago

A 0.50-kg mass attached to the end of a string swings in a vertical circle (radius 2.0 m). When the mass is at the highest point

il63 [147K]

Answer:

31.1 N

Explanation:

m = mass attached to string = 0.50 kg

r = radius of the vertical circle = 2.0 m

v = speed of the mass at the highest point = 12 m/s

T = force of the string on the mass attached.

At the highest point, force equation is given as

$T + mg =\frac{mv^{2}}{r}$

Inserting the values

$T + (0.50)(9.8) =\frac{(0.50)(12)^{2}}{2}$

T = 31.1 N

7 0

2 years ago

Read 2 more answers

A 6.0-ohm resistor that obeys Ohm’s Law is connected to a source of variable potential difference. When the applied voltage is d

leva [86]

Is halved. A 6Ω resistor connected to a voltage source which voltage is decreased from 12V to 6V the current passing through the resistor is halved.

The key to solve this problem is applying Ohm's Law V = R I, clearing I from the equation, we obtain I = V/R. Then, the current is directly proportional to the voltage and inversely proportional to the resistance.

V = 12V and R = 6Ω

I = 12V/6Ω = 2A

V = 6V and R = 6Ω

V = 6V/6Ω = 1A

As we can see the current is halved if the voltage descreased from 12V to 6V

5 0

2 years ago

In a photoelectric effect experiment, electromagnetic radiation containing a finite distribution of wavelengths shines on a meta

Sonja [21]

Answer with Explanation:

a.Intensity of radiation is directly proportional to the frequency of radiation

When the intensity of radiation increases then the frequency of radiation increases and therefore, the number of photo-electrons emitted by the metal increases.

b.When all of the wavelength in the radiation are increased by the same amount

We know that

$f=\frac{v}{\lambda}$

Frequency is inversely proportional to the wavelength.

Therefore, the frequency decrease .

When the frequency decreases then the number of photo-electrons emitted by the metal decrease.

c.When the work function of the metal is increased then the gain of kinetic energy decreases .

When energy decreases then the number of photo-electrons emitted by the metal decreases.

3 0

2 years ago

Determine the correlation between coronal mass ejections from the Sun to the accumulation of the rare and valuable isotope He3 t

Reil [10]

Most ejections originate from active regions on the Sun's surface, such as groupings of sunspots associated with frequent flares. These regions have closed magnetic field lines, in which the magnetic field strength is large enough to contain the plasma.

6 0

2 years ago