The stretchy silk of a certain species of spider has a force constant of 0.80mN/

Question

The stretchy silk of a certain species of spider has a force constant of 0.80mN/cm . The spider, whose mass is 13mg , has attached herself to a branch as shown in the figure

Part A

Calculate the tension in each of the three strands of silk.

Express your answers using two significant figures separated by commas.

Part B

Calculate the distance each strand is stretched beyond its normal length.

Express your answers using two significant figures separated by commas.

The stretchy silk of a certain species of spider has a force constant of 0.80mN/cm . The spider, whose mass is 13mg , has attached herself to a branch as shown in the figure Part A Calculate the tension in each of the three strands of silk. Express your answers using two significant figures separated by commas. Part B Calculate the distance each strand is stretched beyond its normal length. Express your answers using two significant figures separated by commas.

Explanation / Answer

We consider the point at which the three strands intersect. Resolve each of the resultant forces at this point. We know that the tension in strand C is the one borne with the weight of the spider = 13mg X 10 (gravitational constant)=0.13 mN

This vertical downward tension would need to be supported by the vertical components of the strands A & B. Also the horizontal resultant of all these forces is 0. We would thus obtain the following two equations:

A cos(30) + B cos(70)= 0.13 eq(I)

Also A sin(30) = B sin(70) eq(II)

From eq(II), we have A = 1.88B, substituting in eq(I)

1.88B (0.866) + B (0.342) =0.13

B=0.066 mN

Also, from eq(II) A= 0.124 mN

Also since the stretching constant is 0.80 mN/cm, B has stretched by 0.066/0.80=0.0825 cm

Also, A has stretched by 0.124/0.80= 0.155 cm

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