Question

Determine the tension in each wire and the magnitude of each tension.

Short answer

The weight of chain is\(30.1\;{\rm{N}}\).

Step-by-step solution

Formula used

Write the expression for tension vector in terms of horizontal and vertical components (T).

\({\bf{T}} = |{\bf{T}}|\cos \theta {\bf{i}} + |{\bf{T}}|\sin \theta {\bf{j}} \ldots \ldots \ldots (1)\)

Here,

\(|{\bf{T}}|\)is magnitude of tension, and

\(\theta \)is angle of tension.

The weight of chain is counterbalanced by the combined tension contributed by tension at ends of chain.

Write the expression for weight of chain (w)

\(|{\bf{w}}|{\bf{j}} = {{\bf{T}}_1} + {{\bf{T}}_2} \ldots \ldots \ldots (2)\)

Here,

\(|{\bf{w}}|\)is magnitude of weight,

\({{\bf{T}}_1}\)is tension vector of first end of chain, and

\({{\bf{T}}_2}\)is tension vector of second end of chain.

Determine the tension in each wire and the magnitude of each tension

As given, the magnitude of chain at each end as \(25\;{\rm{N}}\) and angle at each end is\({37^^\circ }\).

Draw the vector diagram of chain as shown in figure1

Figure1

Find the tension vector at first end of chain by using equation (1)

\(\begin{aligned}{l}{{\bf{T}}_1} = - (25\;{\rm{N}})\cos \left( {{{37}^^\circ }} \right){\bf{i}} + (25\;{\rm{N}})\sin \left( {{{37}^^\circ }} \right){\bf{j}}\\{{\bf{T}}_1} = - (25)(0.7896){\bf{i}} + (25)(0.6018){\bf{j}}\\{{\bf{T}}_1} = - 19.740{\bf{i}} + 15.045{\bf{j}}\end{aligned}\)

Find the tension vector at second end of chain by using equation (1)\(\begin{aligned}{l}{{\bf{T}}_2} = (25\;{\rm{N}})\cos \left( {{{37}^^\circ }} \right){\bf{i}} + (25\;{\rm{N}})\sin \left( {{{37}^^\circ }} \right){\bf{j}}\\{{\bf{T}}_2} = (25)(0.7896){\bf{i}} + (25)(0.6018){\bf{j}}{\rm{N}}\\{{\bf{T}}_2} = 19.740{\bf{i}} + 15.045{\bf{j}}{\rm{N}}\end{aligned}\)

Substitute \( - 19.740{\bf{i}} + 15.045{\bf{jN}}\) for \({{\bf{T}}_1}\) and \(19.740{\bf{i}} + 15.045{\bf{jN}}\) for \({{\bf{T}}_2}\) in equation (2), \(\begin{aligned}{l}|{\bf{w}}|{\bf{j}} = ( - 19.740{\bf{i}} + 15.045{\bf{j}}){\rm{N}} + (19.740{\bf{i}} + 15.045{\bf{j}}){\rm{N}}\\|{\bf{w}}| = 30.09{\bf{j}}{\rm{N}}\\|{\bf{w}}| \simeq 30.1{\bf{j}}{\rm{N}}\\|{\bf{w}}| = 30.1\;{\rm{N}}\end{aligned}\)

Thus, the weight of chain is\(30.1\;{\rm{N}}\).