Physics, 13.11.2019 04:31 maxy7347go

Auniform stationary ladder of length l and mass m leans against a smooth vertical wall, while its bottom legs rest on a rough horizontal floor. the coefficient of static friction between floor and ladder is μ. the ladder makes an angle θ with respect to the floor. a painter of weight 1/2m stands on the ladder a distance d from its base. find an expression for the magnitude of the normal force n exerted by the floor on the ladder. find an expression for the magnitude of the normal force n_w exerted by the wall on the ladder. find an expression for the largest value of d_max for which the ladder does not slip. what is the largest value for d in centimeters, such that the ladder will not slip? assume that ladder is 3.5 m long, the coefficient of friction is 0.53, the ladder is at an angle of 47.5 degree, and the ladder has a mass of 95 kg.

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Answer from: emfranco1

The expression for the magnitude of the normal force N exerted by the floor on the ladder is $N= \dfrac{3}{2}W$ .

The expression for the magnitude of the normal force $(N_{w})$ exerted by the wall on the ladder is, $N_{w}=\dfrac{3}{2}W \mu$ .

The maximum distance to avoid slipping of ladder is 1.74 m.

Explanation:

Given data:

Weight of painter is, $W=\dfrac{1}{2}M$ .

Length of ladder is, $L = 3.5 \;\rm m$ .

Coefficient of friction is, $\mu = 0.53$ .

Angle of inclination of ladder is, $\theta = 47.5^{\circ}$ .

Mass of ladder is, $M = 95 \;\rm kg$ .

(a)

The expression for the magnitude of the normal force N exerted by the floor on the ladder by translation equilibrium is,

$N-W-\dfrac{W}{2} =0\\N=W+\dfrac{W}{2}\\N=\dfrac{3}{2}W$

Thus, the expression for the magnitude of the normal force N exerted by the floor on the ladder by translation equilibrium is $N= \dfrac{3}{2}W$ . And its value is,

$N= \dfrac{3}{2} \times (\dfrac{M}{2}) \\N= \dfrac{3}{2} \times (\dfrac{95}{2}) \\N = 71.25 \;\rm N$

(b)

The expression for the magnitude of the normal force $(N_{w})$ exerted by the wall on the ladder is,

Since, the normal force exerted by wall on ladder is opposed by the frictional force. Then,

Normal force by wall on ladder = Frictional force

$N_{w}=f\\N_{w}=\mu \times N\\N_{w}=\mu \times (\dfrac{3}{2}W )\\N_{w}=\dfrac{3}{2}W \mu$

Thus, the expression for the magnitude of the normal force $(N_{w})$ exerted by the wall on the ladder is, $N_{w}=\dfrac{3}{2}W \mu$ . And its value is,

$N_{w}=\mu \times N\\N_{w}=0.53 \times 71.25\\N_{w}=37.76 \;\rm N$

(c)

Apply the condition of rotational balance as,

$W \times \dfrac{x}{2} +\dfrac{W}{2}(x-d)+(f y)-Nx=0$

Here,

x is the horizontal distance from the stairs and base of wall.

y is the distance vertical from painter and base of wall.

d is the maximum distance to resist slipping of ladder.

And values are

$y=Lsin \theta\\y=3.5 \times sin47.5 = 2.58 \;\rm m\\x=Lcos \theta\\y=3.5 \times cos47.5 = 2.36 \;\rm m$

Solving as,

$W \times \dfrac{x}{2} +\dfrac{W}{2}(x-d)+(f y)-Nx=0\\\dfrac{M}{2} \times \dfrac{2.36}{2} +\dfrac{M}{4}(2.36-d)+(\mu \times N \times y)-Nx=0\\\dfrac{95}{2} \times \dfrac{2.36}{2} +\dfrac{95}{4}(2.36-d)+(0.53 \times 71.25 \times 2.58)-(71.25 \times 2.36)=0\\\dfrac{95}{4}(2.36-d)=14.67\\d = 1.74 \;\rm m$

Thus, the maximum distance to avoid slipping of ladder is 1.74 m.

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Answer from: miyah5316

a) N = 1396.5 N , b) N_w = 740.1 N c) d = 1.74 m

Explanation

To solve this problem let's use the translational and rotational equilibrium equations

Σ F = 0

Σ τ = 0

We establish an xy reference system, at the highest point of the ladder, where the x-axis parallel to the ground and the y-axis is perpendicular and passes through the vertical wall

Let's write the transnational equilibrium equations

X axis

fr - N_w = 0

Y Axis

N - W - W / 2 = 0

fr = N_w

a) N = 3/2 W

N = 3/2 mg

N = 3/2 95 9.8

N = 1396.5 N

b) fr = N_w

The force of friction is

fr = μ N

fr = μ (3/2 W)

N_w = 3/2 μ m g

N_w = 3/2 0.53 95 9.8

N_w = 740.1 N

c) we write the rotational balance, the anti-hourly rotation is positive

The distance they give to the painter is from the base of the stairs, to measure this distance from the wall is (x-d)

W x / 2 + (W / 2) (x-d) + fr y - N x = 0

Let's use trigonometry to find the distances

sin 47.5 = y / L

y = L sin47.5

cos 47.5 = x / L

x = L cos 47.5

y = 3.5 sin 47.5

x = 3.5 cos 47.5

y = 2.58 m

x = 2.36 m

Let's simplify and calculate d

W x / 2 + (W / 2) (x-d) + fr y - N x = 0

W x + fr y - N x = W / 2 d y -

mg / 2 d = mg x + (μ 3/2 mg) y -3/2 mg x

d = 2 (x + 3/2 μ y - 3/2 x)

d = 2x + 3 (μ L sin 47.5 - L cos 47.5)

d = 2L cos 47.5 + 3L (μ sin 47.5 - cos 47.5)

d = L cos 47.5 (2 + 3 (miu tan 47.5 -1))

d = 3.5 cos 47.5 (2 + 3 (0.53 tan47.5 -1))

d = 2.3645 (2 - 1.2648)

d = 1.74 m

Answer from: Quest

answer: the direction of the wind is expressed as the direction from which the wind is blowing. for example, easterly winds blow from east to west, while westerly winds blow from west to east.

explanation:

Answer from: Quest

The answer is a. i hope this

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