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

Two wires are stretched between two fixed supports and have the same length. One wire A there is a second-harmonic standing wave

lina2011 [118]

(a) Greater

The frequency of the nth-harmonic on a string is an integer multiple of the fundamental frequency, $f_1$ :

$f_n = n f_1$

So we have:

- On wire A, the second-harmonic has frequency of $f_2 = 660 Hz$ , so the fundamental frequency is:

$f_1 = \frac{f_2}{2}=\frac{660 Hz}{2}=330 Hz$

- On wire B, the third-harmonic has frequency of $f_3 = 660 Hz$ , so the fundamental frequency is

$f_1 = \frac{f_3}{3}=\frac{660 Hz}{3}=220 Hz$

So, the fundamental frequency of wire A is greater than the fundamental frequency of wire B.

(b) $f_1 = \frac{v}{2L}$

For standing waves on a string, the fundamental frequency is given by the formula:

$f_1 = \frac{v}{2L}$

where

v is the speed at which the waves travel back and forth on the wire

L is the length of the string

(c) Greater speed on wire A

We can solve the formula of the fundamental frequency for v, the speed of the wave:

$v=2Lf_1$

We know that the two wires have same length L. For wire A, $f_1 = 330 Hz$ , while for wave B, $f_B = 220 Hz$ , so we can write the ratio between the speeds of the waves in the two wires:

$\frac{v_A}{v_B}=\frac{2L(330 Hz)}{2L(220 Hz)}=\frac{3}{2}$

So, the waves travel faster on wire A.

7 0

2 years ago

A horizontal spring with spring constant 85 n/m extends outward from a wall just above floor level. a 3.5 kg box sliding across

Rina8888 [55]

k = spring constant of the spring = 85 N/m

m = mass of the box sliding towards the spring = 3.5 kg

v = speed of box just before colliding with the spring = ?

x = compression the spring = 6.5 cm = 6.5 cm (1 m /100 cm) = 0.065 m

the kinetic energy of box just before colliding with the spring converts into the spring energy of the spring when it is fully compressed.

Using conservation of energy

Kinetic energy of spring before collision = spring energy of spring after compression

(0.5) m v² = (0.5) k x²

m v² = k x²

inserting the values

(3.5 kg) v² = (85 N/m) (0.065 m)²

v = 0.32 m/s

8 0

2 years ago

14 gauge copper wire has a diameter of 1.6 mm. what length of this wire has a resistance of 4.8ω?

Vladimir79 [104]

The relationship between resistance R and resistivity $\rho$ is
$R= \frac{\rho L}{A}$
where L is the length of the wire and A its cross section.

The radius of the wire is half the diameter:
$r= \frac{d}{2}= \frac{1.6 mm}{2}=0.8 mm=8\cdot 10^{-4} m$
and the cross section is
$A=\pi r^2 = \pi (8\cdot 10^{-4} m)^2=2.01\cdot 10^{-6} m^2$

From the first equation, we can then find the length of the wire when $R=4.8 \Omega$ (copper resistivity: $\rho = 1.724 \cdot 10^{-8} \Omega m$ )
$L= \frac{AR}{\rho}= \frac{(2.01\cdot 10^{-6} m^2)(1.724 \cdot 10^{-8} \Omega m)}{4.8 \Omega}=7.21 \cdot 10^{-15} m$

4 0

2 years ago

A uniform magnetic field of 0.50 T is directed along the positive x axis. A proton moving with a speed of 60 km s enters this fi

tatuchka [14]

Explanation:

It is given that,

Magnetic field, B = 0.5 T

Speed of the proton, v = 60 km/s = 60000 m/s

The helical path followed by the proton shown has a pitch of 5.0 mm, p = 0.005 m

We need to find the angle between the magnetic field and the velocity of the proton. The pitch of the helix is the product of parallel component of velocity and time period. Mathematically, it is given by :

$p=v_{||}\times T$

$p=v\ cos\theta\times \dfrac{2\pi m}{Bq}$

$cos\theta=\dfrac{pBq}{2\pi mv}$

$cos\theta=\dfrac{0.005\times 0.5\times 1.6\times 10^{-19}}{2\pi \times 1.67\times 10^{-27}\times 60000}$

$\theta=50.58^{\circ}$

So, the angle between the magnetic field and the velocity of the proton is 50.58 degrees. Hence, this is the required solution.

3 0

2 years ago

The equation for the change in the position of a train (measured in units of length) is given by the following expression: x = ½

mel-nik [20]

Answer:

(B) (length)/(time³)

Explanation

The equation x = ½ at² + bt³ has to be dimensionally correct. In other words the term bt³ and ½ at² must have units of change of position = length.

We solve in order to find the dimension of b:

[x]=[b]*[t]³

length=[b]*time³

[b]=length/time³

6 0

2 years ago

Read 2 more answers