Find the force required to move a load of 255 N up a rough inclined plane, the force being applied parallel to the plane. The inclination of the plane is such that when the same body is kept on a perfectly smooth plane inclined at that angle, a force of 55 N applied at an inclination of 45° to the plane keeps the same in equilibrium. Assume that co-efficient of friction between the rough plane and load is equal to 0.35.
Answers
Answer:
ANSWER
Let the coefficient of friction be μ
Thus friction force acting f=μmg cos60
o
=
2
μmg
Upward motion : Friction force acts in downward direction
∴ F
u
=mgsin60
o
+f=
2
3
mg+
2
μmg
=mg[
2
3
+μ
]
Downward motion : Friction force acts in upward direction
∴ F
d
=mgsin60
o
−f=
2
3
mg−
2
μmg
=mg[
2
3
−μ
]
According to the question, F
u
=2F
d
mg[
2
3
+μ
]= 2×mg[
2
3
−μ
]
OR
3
+μ=2[
3
−μ] ⟹μ=
3
1
solution
Explanation:
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In the Isosceles Triangle ABC,
∠BAC = 30°
So,
∠CBA also = 30° = x (Opposite side in a Isosceles ∆)
_____________________________________
If ∠BAC & ∠CBA = 30°
Then,
∠BAC + ∠CBA + ∠ACB = 180° (Angle Sum Prop. of a Triangle)
➠ 30° + 30° + ∠ACB = 180°
➠ ∠ACB = 180° - 60°
➠ ∠ACB = 120°
If ACB = 120° Then,
∠BCD = 180° - 120°
∠BCD = z = 60°
___________________________________
If z = 60°
Then,
Y also = 60° (Opp. Sides in an Isosceles ∆ )
____________________________________
Overall,
x = 30°
y = 60°
z = 60°