A ladder of length 5 m and weight 300 N is placed against a vertical wall
with which it makes an angle of 45o
. The coefficient of friction between the floor
and the ladder is 0.5 and that between the wall and the ladder is 0.4. In addition
to its own weight, the ladder has to support a man of weight 500 N at 1m from
the top along the ladder. Determine the minimum horizontal force ‘P’ to be
applied at the floor level to prevent the ladder from slipping
Answers
Answer:
Answer
Correct option is
C
μ=1/3
Let mass of ladder be m. consider free body diagram of ladder.
For verticle equilibrium -
N
1
+μN
2
=mg __(1)
For Horizontal equilibrium
N
2
=μN
1
__(2)
Torqul about point A
mg(2.5cosθ)=N
2
(4)+(μN
2
)
3
mg(2.5×
5
3
)=(4+3μ)N
2
(cosθ=
5
3
)
mg(
2
3
)=(4+3μ)N
2
___(3)
From (1) and (2) N
2
[
μ
1+μ
2
]=mg⇒N
2
=
1+μ
2
μmg
Hence mg(
2
3
)=(4+3μ)[
1+μ
2
μmg
]
(3+3μ
2
)=8μ+6μ
2
3μ
2
+6μ−3=0
⇒μ=
6
−8±
64+36
=
6
−8±10
μ=
3
1
Hence, option (C) is correct answer.
solution
Answer:
P = 850.9 N
what is friction?
Friction is a force that opposes motion between two surfaces in contact with each other.
Explanation:
To prevent the ladder from slipping, the forces acting on it must be in equilibrium. Let's consider the forces acting on the ladder:
Weight of the ladder: 300 N acting downwards from the center of mass.
Weight of the man: 500 N acting downwards 1m from the top.
Reaction force from the wall: R_w acting perpendicular to the wall.
Frictional force from the wall: f_w = 0.4 R_w acting parallel to the wall.
Reaction force from the floor: R_f acting perpendicular to the floor.
Frictional force from the floor: f_f = 0.5 R_f acting parallel to the floor.
R_f + R_w - 300 N - 500 N = 0
R_f + R_w = 800 N
Next, let's consider the forces in the horizontal direction:
P - f_f - f_w = 0
P = f_f + f_w
The frictional forces can be calculated using the coefficients of friction:
f_f = 0.5 R_f
f_w = 0.4 R_w
Substituting these equations into the equation for P, we get:
P = 0.5 R_f + 0.4 R_w
To solve for P, we need to eliminate R_f and R_w. The angle between the ladder and the floor is 45 degrees, so we can write:
sin 45 = R_w / R_f
Solving for R_w in terms of R_f, we get:
R_w = R_f / sqrt(2)
Substituting this equation into the equation for P, we get:
P = 0.5 R_f + 0.4 (R_f / sqrt(2))
Simplifying, we get:
P = R_f (0.5 + 0.4 / sqrt(2))
Substituting the equation for R_f from the equation for vertical forces, we get:
P = 800 N (0.5 + 0.4 / sqrt(2))
Simplifying, we get:
P = 850.9 N
To learn more about friction follow the given link :
https://brainly.in/question/46756874
https://brainly.in/question/1539421
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