Ask question

Ask question

All categories

valentina_108 [34]

3 years ago

6

A 5.00-m-long uniform ladder, weighing 200 N, rests against a smooth vertical wall with its base on a horizontal rough floor, a

distance of 1.20 m away from the wall. The coefficient of static friction between the ladder and the floor is 0.200. How far up the ladder, measured along the ladder, can a 600-N person climb before the ladder begins to slip

1 answer:

Morgarella [4.7K]3 years ago

5 0

Answer:

0.488 m

Explanation:

If θ be the angle ladder makes with the plane

cos θ = 1.2 / 5

Tan θ = 4.04

Let the height a person of weight 600 N can climb be h from the ground .

Distance from the base point where ladder touches the floor = h / tanθ

= h / 4.04

Total reaction force = total downward force

R = 200 + 600

800 N

Frictional force = μ R

= .2 x 800

= 160 N

Taking moment of force about the point on the ladder where it touches the floor and balancing them

200 x 1.2 x .5 + 600 x h / tanθ = μ R x 1.2 / tanθ ( reaction at the top point of ladder where it touches the wall is R₁ and

R₁ =μ R )

= 200 x 1.2 x .5 + 600 x h / tanθ = 160 x 1.2 / tanθ

120 - 600 h / 4.04 = 47.52

120 - 47.52 = 600 h / 4.04

72.48= 148.51 h

h = 0.488 m

=

You might be interested in

A string is stretched and fixed at both ends, 200 cm apart. If the density of the string is 0.015 g/cm, and its tension is 600 N

Gre4nikov [31]

Answer:

<h2>f₀ = 158.12 Hertz</h2>

Explanation:

The fundamental frequency of the string f₀ is expressed as f₀ = V/4L where V is the speed experienced by the string.

$V = \sqrt{\frac{T}{\mu} }$ where T is the tension in the string and $\mu$ is the density of the string

Given T = 600N and $\mu$ = 0.015 g/cm = 0.0015kg/m

$V = \sqrt{\frac{600}{0.0015} }\\ \\V = \sqrt{400,000}\\ \\V = 632.46m/s$

The next is to get the length L of the string. Since the string is stretched and fixed at both ends, 200 cm apart, then the length of the string in metres is 2m.

L = 2m

Substituting the derived values into the formula f₀ = V/2L

f₀ = 632.46/2(2)

f₀ = 632.46/4

f₀ = 158.12 Hertz

Hence the fundamental frequency of the string is 158.12 Hertz

5 0

3 years ago

Which type of bond forms when two or more atoms share electrons?

Oxana [17]

Covalent bonds form when two or more atoms share electrons

3 0

3 years ago

Two cars A and B, travel in a straight line. The distance of A from the starting point is given as a function of time by x????(?

Norma-Jean [14]

Answer:

a) They are in the same point

b) t = 0 s, t = 2.27 s, t = 5.73 s

c) t = 1 s, t = 4.33 s

d) t = 2.67 s

Explanation:

Given equations are:

$x_{a}(t) = at+bt^2$

$x_{b}(t) = ct^2-dt^3$

Constants are:

$a = 2.60 m/s, b = 1.20 m/s^2, c= 2.80 m/s^2, d = 0.20 m/s^3$

a) "Just after leaving the starting point" means that t = 0. So, if we look the equations, both $x_a(t)$ and $x_b(t)$ depend on t and don't have constant terms.

So both cars A and B are in the same point.

b) Firstly, they are in the same point in x = 0 at t = 0. But for generalized case, we must equalize equations and solve quadratic equation where roots will give us proper t value(s).

$at+bt^2 = ct^2-dt^3$

$2.6t + 1.2t^2 = 2.8t^2 - 0.2t^3\\0.2t^2 - 1.6t + 2.6 = 0\\t^2 - 8t + 13 = 0$

$t_1 = 4 - \sqrt{3} = 2.27$ s, $t_1 = 4 + \sqrt{3} = 5.73$ s

c) Since the distance isn't changing, the velocities are equal. To find velocities, we need to take the derivatives of both equations with respect to time and equalize them.

$v_a(t) = \frac{d}{d(t)}x_a(t) = a + 2bt \\v_b(t) = \frac{d}{d(t)}x_b(t) = 2ct - 3dt^2\\a+2bt = 2ct - 3dt^2\\3dt^2+2(b-c)t+a = 0\\0.6t^2-3.2t+2.6 = 0$

$t_1 = 1$ s, $t_2 = 4.33$ s

d) For same acceleration, we we need to take the derivatives of velocity equations with respect to time and equalize them.

$a_a(t) = \frac{d}{d(t)}v_a(t) = 2b \\a_b(t) = \frac{d}{d(t)}v_b(t) = 2c - 6dt\\2b = 2c - 6dt\\3dt = c - b\\t = (c - b)/3d = (2.8 - 1.2)/(3*0.2) = 2.67$ s

3 0

3 years ago

Read 2 more answers

A series circuit that is connected to a 50 V, 60 Hz source is made up of 25 ohm resistor, capacite wieh X= 18 ohms, and inductor

Allisa [31]

Answer:

the impedance of the circuit is 25.7 ohms.

Explanation:

It is given that,

Voltage, V = 50 volts

Frequency, f = 60 Hz

Resistance, R = 25 ohms

Capacitive resistance, $X_C=18\ ohms$

Inductive resistance, $X_L=24\ ohms$

We need to find the impedance of the circuit. It is given by :

$Z=\sqrt{R^2+(X_L-X_C)^2}$

$Z=\sqrt{25^2+(24-18)^2}$

Z = 25.7 ohms

So, the impedance of the circuit is 25.7 ohms. Hence, this is the required solution.

6 0

3 years ago

Give three examples from your life of magnetic force.

Margaret [11]

Headphones, refrigerator magnets, and compasses

Hope that was helpful.

4 0

3 years ago

Read 2 more answers