Consider an advertising board of mass ‘m’ hangs with the help of two strings making equal
angles with the ceiling. Calculate the tension in both the strings??
Answers
Answer:
Solution: The free body diagram of the same helps us to resolve the forces first. After resolving the forces we will apply the required theorem to get the value of tension in both the strings. Here, the weight of the signboard is in a downward direction, and the other force is the tension generated by the signboard in both the strings. In this case, the tension T in both the strings will be the same as the angle made by them with the signboard is equal.
Free body diagram of the signboard in the. attachment .
Above figure represents the free body diagram of the signboard. Applying the Lami’s Theorem we get,
T/sin (180- θ)= T/ sin 180-θ =mg/sin 2 θ
Since sin (180 – θ) = sin θ and sin (2θ) = 2sinθ cosθ
So, we get, final tension force in the string T as,T/sin θ = mg/ 2 sin θ × cos θ I.e. T = mg / 2cosθ
The similar concept and equations can be applied for a boy playing on a swing, and we arrive at the same result.
Answer:
\huge{\sf{Question}}Question
Two sides of a triangle are 5 and 7 cm longer than the third side. If the perimeter is 21cm, find the length of each side ?
\huge{\sf{Given:-}}Given:−
Let the third side be x
Then,
Length of first side = (x+5)cm
Length of second side = (x+7)cm
Length of third side = x cm
Perimeter = 21cm
\huge{\sf{Solution:-}}Solution:−
Perimeter of triangle = Sum of their sides
\longrightarrow⟶ \sf{(x+5+x+7+x)cm= 21cm}(x+5+x+7+x)cm=21cm
\longrightarrow⟶ \sf{(3x+12)cm = 21cm}(3x+12)cm=21cm
\longrightarrow⟶ \sf{3x=(21-12)cm}3x=(21−12)cm
\longrightarrow⟶ \sf{3x=9}3x=9
\longrightarrow⟶ \sf{x= \frac{9}{3}}x=
3
9
\longrightarrow⟶ \sf{x=3}x=3
_______________________________
\sf{\red{Therefore,the\: sides\:are:-}}Therefore,thesidesare:−
First side = (3+5)=8cm
Second side = (3+7)=10cm
Third side = 3cm