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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answered: 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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answered: 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