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4-1.Newton's Laws of Motion
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A $100$ $Newton$ weight is suspended in a corner of a room by two cords $A$ and $B$ as shown in the figure below. The tension in the horizontal cord $A$ is ............ $N$
A
$50$
B
$100$
C
$200$
D
$173.2$
Solution
$\mathrm{T}_{\mathrm{B}} \cos 60^{\circ}=100$
$\mathrm{T}_{\mathrm{B}}=200$
$\mathrm{T}_{\mathrm{B}} \sin 60^{\circ}=\mathrm{T}_{\mathrm{A}}$
$\mathrm{T}_{\mathrm{A}}=100 \sqrt{3}=173.2 \mathrm{N}$
Standard 11
Physics
