(II) A 0.75-kg sheet is centered on a clothesline as shown in Fig. 9–63. The clothesline on either side of the hanging sheet makes an angle of 3.5° with the horizontal. Calculate the tension in the clothesline (ignore its mass) on either side of the sheet. Why is the tension so much greater than the weight of the sheet?

Translational equilibrium may be described as the state in which an object is at rest or in un-accelerated motion, where the resultant of all forces working on it is zero.

Given data:

The mass of the sheet is\(m = 0.75\;{\rm{kg}}\).

The angle of inclination is \(\theta = 3.5^\circ \).

The free-body diagram for the sheet can be drawn as:

Here, \({F_{\rm{T}}}\) is the tension force on each side.

Now, apply the force equilibrium condition along the vertical direction.

\(\begin{array}{c}\sum {F_{\rm{y}}} = 0\\{F_{\rm{T}}}\sin \theta + {F_{\rm{T}}}\sin \theta - mg = 0\\{F_{\rm{T}}}\sin \left( {3.5^\circ } \right) + {F_{\rm{T}}}\sin \left( {3.5^\circ } \right) - \left( {0.75\;{\rm{kg}}} \right)\left( {9.8\;{{\rm{m}} \mathord{\left/{\vphantom {{\rm{m}} {{{\rm{s}}^{\rm{2}}}}}} \right.} {{{\rm{s}}^{\rm{2}}}}}} \right) = 0\\{F_{\rm{T}}} = 60.1\;{\rm{N}} \approx 60\;{\rm{N}}\end{array}\)

Thus, the tension in the clothesline on either side of the sheet is \(60\;{\rm{N}}\).

So, according to the result, \(60\;{\rm{N}}\) tension is higher than the \({\rm{7}}{\rm{.35}}\;{\rm{N}}\) weight because of the small angle.

Only the vertical components of the tension support the sheet. Since the angle is small, the tension must be greater to have a large enough vertical component to keep the sheet up.