A body, weighing 10kg is suspended by two strings AC and BC at the point C .The length of the string AC=3m and BC =4m respectively. the horizontal distance of BC is 4 meters. find the tensions in the string AC and BC
Answers
Answered by
0
Answer:
Correct option is
C
5 kgf and 12 kgf
From the figure,
sinθ=
13
5
cosθ=
13
12
The horizontal force on the body balance each other.
Thus T
1
cosθ=T
2
cos(
2
π
−θ)
⟹T
1
cosθ=T
2
sinθ
Also the vertical components balance each other,
⟹13kgf=T
2
cosθ+T
1
sin(
2
π
−θ)
⟹13kgf=T
2
cosθ+T
1
sinθ
⟹T
1
=5kgf
T
2
=12kgf
Answered by
0
Answer:
Since the system is in equilibrium;
The moment of forces about point A should be zero.
Thus;
10g×0.5+15g×1=Tsin53
o
×1
50+150=0.8T
T=250 N
At point A, the hinge force is working which can be resolved as R
y
& R
x
R
x
=Tcos 53
o
R
x
=250×0.6=150 N
R
y
=Tsin53
o
R
y
=250×0.8=200 N
R=
(R
x
)
2
+(R
y
)
2
R=
150
2
+200
2
R=250 N
Therefore, The tension in the string will be 250 N and the hinged force at point A will be 250 N
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