A uniform ladder of length 10.0 m and mass 10.0 kg is resting against a vertical wall making an angle of 37° with it. The vertical wall is frictionless but the ground is rough. An electrician weighing 60.0 kg climbs up the ladder. If he stays on the ladder at a point 8.00 m from the 53x lower end, the frictional force on the ladder by the ground is 53x/3. find the value of x. (g = 10 m/s )
Good Morning
Answers
Step-by-step explanation:
Solution
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Let the forces F
1
and F
2
as shown in the above figure.
Since, the ladder should not slip or rotate.
The torques expression is given as,
mg(8sin37
∘
)+Mg(5sin37
∘
)−F
2
(10cos37
∘
)=0
60×9.8×(8sin37
∘
)+16×9.8×(5sin37
∘
)−F
2
(10cos37
∘
)=0
F
2
=412N
Since, from the equilibrium of the ladder the force F
2
will be equal to the friction force.
The friction force is given as,
f=412N
The normal force is given as,
F
1
=(m+M)g
=(60+16)×9.8
=744.8N
The minimum coefficient of friction is given as,
μ=
F
1
f
=
744.8
412
=0.553
Thus, the minimum coefficient of friction is 0.553
Good morning
Answer:
Let the forces F
1
and F
2
as shown in the above figure.
Since, the ladder should not slip or rotate.
The torques expression is given as,
mg(8sin37
∘
)+Mg(5sin37
∘
)−F
2
(10cos37
∘
)=0
60×9.8×(8sin37
∘
)+16×9.8×(5sin37
∘
)−F
2
(10cos37
∘
)=0
F
2
=412N
Since, from the equilibrium of the ladder the force F
2
will be equal to the friction force.
The friction force is given as,
f=412N
The normal force is given as,
F
1
=(m+M)g
=(60+16)×9.8
=744.8N
The minimum coefficient of friction is given as,
μ=
F
1
f
=
744.8
412
=0.553
Thus, the minimum coefficient of friction is 0.553.