Q19PE - College Physics (Urone)

Question

(a) Calculate the tension in a vertical strand of spider web if a spider of mass 8.00×10⁻⁵ kg hangs motionless on it. 

(b) Calculate the tension in a horizontal strand of spider web if the same spider sits motionless in the middle of it, much like the tightrope walker in Figure 4.17. The strand sags at an angle of 12º below the horizontal. Compare this with the tension in the vertical strand (find their ratio).

Step-by-Step Solution

Answer

(a) The tension in the vertical strand is 7.84 x 10^-4 N.

(b) The tension in the horizontal strand is 1.89 x 10^-3 N. The tension in the horizontal strand is 2.4 times the tension in the vertical strand.

Step 1: Concept of Normal Force.

Whatever supports a load, it must supply an upward force equal to the weight of the load. This upward force is called the normal force.

 

Step 2: (a) Determine the tension in the vertical strand of a spider web

The tension which acts upwards must be equal to the weight of the spider can be written as,

 

\(T = mg\)

 

Here, T is the tension in the vertical strand, m is the mass of the spider, and g is the acceleration due to gravity.

 

Substitute 8x10^-5 kg for m and 9.8 m/s² for g in the above expression, and we get,

 

\(\begin{array}{l}T = 8 \times {10^{ - 5}}\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}\\T = 7.84 \times {10^{ - 4}}\;{\rm{N}}\end{array}\)

 

Hence, the tension in the vertical strand is 7.84x10^-4 N.

 

Step 3: (b) Determine the tension in the horizontal strand of the spider web

Write the expression or calculate the tension in the horizontal strand as:

 

\(T' = \frac{{mg}}{{2\sin \theta }}\)

 

Here, T’ is the tension in the horizontal strand and \(\theta \) is the angle at which the strand sags below the horizontal.

 

Substitute 8x10^-5 kg for m, 12⁰ for \(\theta \) , and 9.8 m/s² for g in the above expression, and we get,

 

\(\begin{array}{c}T' = \frac{{8 \times {{10}^{ - 5}}\;{\rm{kg}} \times 9.8\;{\rm{m/}}{{\rm{s}}^{\rm{2}}}}}{{2 \times \sin {{12}^\circ }}}\\ = \frac{{7.84 \times {{10}^{ - 4}}\;{\rm{N}}}}{{2 \times 0.2079}}\\ = 1.89 \times {10^{ - 3}}\;{\rm{N}}\end{array}\)

 

Hence, the tension in the horizontal strand is 1.89x10^-3 N.

Step 4: Comparison of the tensions in the vertical strand with the horizontal strand of the spider web

\(\begin{array}{c}\frac{{T'}}{T} = \frac{{1.89 \times {{10}^{ - 3}}}}{{7.84 \times {{10}^{ - 4}}}}\\ = 2.4\end{array}\)

 

Hence, the tension in the horizontal strand is 2.4 times more than that in the vertical strand.