A and B are two nails on a horizontal line. AC, BC are two uniform strings from which a weight of 100 N is suspended. The tension of AC is: (in N):

Question

A and B are two nails on a horizontal line. AC, BC are two uniform strings from which a weight of 100 N is suspended. The tension of AC is: (in N): (A) 100 N (B) 80 N (C) 60 N (D) 50 N

Tardigrade Admin · Accepted Answer

According Lamis theorem (T1/ sin 180° -θ )=(T2/ sin (90° +θ ))=(W/ sin 90° ) sin θ =(4/5), cos θ =(3/5) So, T1=W sin θ =100× 4 text/5=80N

Q. A and B are two nails on a horizontal line. AC, BC are two uniform strings from which a weight of 100 N is suspended. The tension of AC is: (in N):

100 N

80 N

60 N

50 N

According Lamis theorem
$\frac{T _{1}}{s i n 180 ^{\circ} - θ} = \frac{T _{2}}{s i n ( 90 ^{\circ} + θ )} = \frac{W}{s i n 90 ^{\circ}}$ $sin θ = \frac{4}{5},$ $cos θ = \frac{3}{5}$ So, $T_{1} = W sin θ = 100 \times 4 / 5 = 80 N$

Q. A and B are two nails on a horizontal line. AC, BC are two uniform strings from which a weight of 100 N is suspended. The tension of AC is: (in N):

100 N

80 N

60 N

50 N

According Lamis theorem sin180∘−θT1​​=sin(90∘+θ)T2​​=sin90∘W​ sinθ=54​, cosθ=53​ So, T1​=Wsinθ=100×4/5=80N

According Lamis theorem
$\frac{T _{1}}{s i n 180 ^{\circ} - θ} = \frac{T _{2}}{s i n ( 90 ^{\circ} + θ )} = \frac{W}{s i n 90 ^{\circ}}$ $sin θ = \frac{4}{5},$ $cos θ = \frac{3}{5}$ So, $T_{1} = W sin θ = 100 \times 4 / 5 = 80 N$