a uniform rod ab of weight 100 N and length 2 m supported by two vertical strings at its ends and carries a load of 20 KGF. find tension in the string
Answers
Answer:
Explanation:
iven: A uniform metre stick having a mass 400g is suspended from the fixed supports through two vertical light strings of equal lengths fixed at the ends. A small object of mass 100g is put on the stick at a distance of 60cm from the left end.
To find the tension in the two strings.
Solution:
Let the tension in the left string be T
1
and the right string be T
2
.
Three forces are acting on the meter stick, i.e., T
1
,T
2
and mg, where m is the mass of the small object. As the rod does not rotate so the net torque is zero and the system is in the translational equilibrium, so we have the condition that the sum of torques around any point on the stick must be zero.
We can choose any point about which to do our torque analysis, let's choose the left end of the stick since the torque due to T
1
will be zero
(T
1
×r
1
)−(T
2
×r
2
)−(m
1
g)(m
2
)=0
As per the given condition,
m
1
=400g=0.4kg,r
1
=60cm=0.6m,r
2
=1−0.6=0.4m,l=1m,m
2
=100g=0.1kg
⟹T
1
×0.6−T
2
×0.4−0.4×0.1×9.8=0
⟹0.6T
1
−0.4T
2
=0.392.......(i)
The two tensions must support the total weight of the system, so we have
T
1
+T
2
=0.4×9.8+0.1×9.8=4.9........(ii)
Multiply eqn(ii) with 0.4 and add with eqn(i), we get
0.6T
1
−0.4T
2
=0.392
+
0.4T
1
+0.4T
2
=1.96
⟹T
1
=2.35N
Substitute the value of T
1
in eqn(ii) we get,
T
1
+T
2
=4.9
⟹2.35+T
2
=4.9
⟹T
2
=2.55N
Hence the tension in the two strings are 2.35N and 2.55N
Answer:
the answer is option c 35 kgf...,