Respuesta :
Answer:
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:
The expression for the magnitude of the normal force N exerted by the floor on the ladder is [tex]N= \dfrac{3}{2}W[/tex].
The expression for the magnitude of the normal force [tex](N_{w})[/tex] exerted by the wall on the ladder is, [tex]N_{w}=\dfrac{3}{2}W \mu[/tex].
The maximum distance to avoid slipping of ladder is 1.74 m.
Explanation:
Given data:
Weight of painter is, [tex]W=\dfrac{1}{2}M[/tex].
Length of ladder is, [tex]L = 3.5 \;\rm m[/tex].
Coefficient of friction is, [tex]\mu = 0.53[/tex].
Angle of inclination of ladder is, [tex]\theta = 47.5^{\circ}[/tex].
Mass of ladder is, [tex]M = 95 \;\rm kg[/tex].
(a)
The expression for the magnitude of the normal force N exerted by the floor on the ladder by translation equilibrium is,
[tex]N-W-\dfrac{W}{2} =0\\N=W+\dfrac{W}{2}\\N=\dfrac{3}{2}W[/tex]
Thus, the expression for the magnitude of the normal force N exerted by the floor on the ladder by translation equilibrium is [tex]N= \dfrac{3}{2}W[/tex]. And its value is,
[tex]N= \dfrac{3}{2} \times (\dfrac{M}{2}) \\N= \dfrac{3}{2} \times (\dfrac{95}{2}) \\N = 71.25 \;\rm N[/tex]
(b)
The expression for the magnitude of the normal force [tex](N_{w})[/tex] 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
[tex]N_{w}=f\\N_{w}=\mu \times N\\N_{w}=\mu \times (\dfrac{3}{2}W )\\N_{w}=\dfrac{3}{2}W \mu[/tex]
Thus, the expression for the magnitude of the normal force [tex](N_{w})[/tex] exerted by the wall on the ladder is, [tex]N_{w}=\dfrac{3}{2}W \mu[/tex]. And its value is,
[tex]N_{w}=\mu \times N\\N_{w}=0.53 \times 71.25\\N_{w}=37.76 \;\rm N[/tex]
(c)
Apply the condition of rotational balance as,
[tex]W \times \dfrac{x}{2} +\dfrac{W}{2}(x-d)+(f y)-Nx=0[/tex]
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
[tex]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[/tex]
Solving as,
[tex]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[/tex]
Thus, the maximum distance to avoid slipping of ladder is 1.74 m.
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