Question

For the following questions, please provide a complete step by step solution. You do notneed interaction, but you are required force diagrams.

Answer 1

Tennsion 1= horizontal = 484.94 N

Tension 2 = slanted string = 560 N

Step-by-step explanation

Step 1

Free body diagram

Newton's first law says that if the net force on an object is zero, like in this case the mass is in rest,then that object will have zero acceleration

so

Step 1

set the equations:

a) for x-axis

`\begin{gathered} \sum ^{}_{}F_x=0 \\ so \\ T_(2x)-T_1=0 \\ T_2\cos 30-T_1=0\rightarrow equation(1) \end{gathered}`

b) for y -axis

`\begin{gathered} \sum ^{}_{}F_y=0 \\ so \\ T_(2y)-w=0 \\ T_2\sin 30-280N=0\rightarrow equation(2) \end{gathered}`

Step 2

Solve the equations

`\begin{gathered} T_2\cos 30-T_1=0\rightarrow equation(1) \\ T_2\sin 30-280N=0\rightarrow equation(2) \end{gathered}`

a) solve for T2in equation (2)

`\begin{gathered} T_2\sin 30-280N=0\rightarrow equation(2 \\ \text{add 280 N in both sides} \\ T_2\sin 30-280N+280N=0+280\text{ N} \\ T_2\sin 30=280\text{ N} \\ \text{divide both sides by sin 30} \\ (T_2\sin30)/(\sin30)=\frac{280\text{ N}}{\sin30} \\ T_2=560\text{ N} \end{gathered}`

b) replace the T2 value in equation (1) to find T1

`\begin{gathered} T_2\cos 30-T_1=0\rightarrow equation(1) \\ 560\cos 30-T_1=0\rightarrow equation(1) \\ 484.97-T_1=0 \\ 484.97=T_1 \end{gathered}`

therefore

Tennsion 1= horizontal = 484.94 N

Tension 2 = slanted string = 560 N

I hope this helps you