Question

If M-16 kg, what is the tension in string 2? The angles shown are 30° for string 1 and 60° for string 2.

Answer 1

Since the block is at rest, i.e., equilibrium, the net force on the system must be zero: That means:

`\begin{gathered} F_(1x)=F_(2x) \\ F_1\sin 30^o=F_2\sin 60^o \\ F_1=(F_2\sin 60^o)/(\sin 30^o)^{}\rightarrow eq(1) \end{gathered}`

Also we have:

`\begin{gathered} F_(1y)+F_(2y)=W \\ F_1\cos 30^o+F_2\cos 60^o=mg \\ (F_2\sin60^o)/(\sin30^o)\cos 30^o+F_2\cos 60^o=mg \\ F_2\sin 60^o\cot 30^o+F_2\cos 60^o=mg \\ F_2(\sin 60^o\cot 30^o+\cos 60^o)=mg \\ F_2=(mg)/(\sin 60^o\cot 30^o+\cos 60^o) \\ \end{gathered}`

Lets say that g = 9.8 m/s², then if we plug the data we get:

`\begin{gathered} F_2=(16*9.8)/((0.87)*(1.73)+(0.50)) \\ \\ F_2=(156.8)/(2.01) \\ \\ F_2=78N \end{gathered}`

Answer: the tension in string 2 is 78 N.