A block mass 1 kg is suspended by a string of mass 1 kg l 1m as shown in figure g 10m/s2 calculate
Answers
Explanation:
T=10N
T
1
=15N
T
2
=20N
Given :-
\begin{lgathered}m = 1 kg \\l = 1 m \\ g = 10 m/s^2\end{lgathered}
m=1kg
l=1m
g=10m/s
2
To find :-
(i) the tension in string at its lowest point.
(ii) the tension in string at its mid-point.
(iii) force Exerted by support on stri
Solution:-
Take block as system.
Force acting on block :-
1) mg force downward
2) Tension force upward.
The system is in equilibrium.
→\mathsf{ T = mg}T=mg
→\mathsf{ T = 1 \times 10}T=1×10
→\mathsf{ T = 10 N}T=10N
hence,
Tension force in string at lowest point is 10 N.
Case :- 2
Tension force at mid - point of the string.
L = m
L/2 = m /2
m = mass of block + mass of half string.
→\mathsf{M = 1kg + 0.5 kg}M=1kg+0.5kg
→\mathsf{ M = 1.5 kg}M=1.5kg
→\mathsf{ T_1 = Mg}T
1
=Mg
→\mathsf{T_1 = 1.5 \times 10}T
1
=1.5×10
→\mathsf{ T_1 = 15 N}T
1
=15N
hence,
Tension force at middle of string is 15 N.
Case :- 3
Force exerted by support on string :-
Tension force by support = Tension force by string + mg force by block.
M" = mass of the block + mass of the string.
M" = 1 + 1
M" = 2 kg
→\mathsf{T_2 = M"g}T
2
=M"g
→\mathsf{T_2 = 2 \times 10}T
2
=2×10
→\mathsf{ T_2 = 20 N}T
2
=20N
hence, tension force exerted by support on string is 20 N.