The stretchy silk of a certain species of spider behaves much like a spring, obe

Question

The stretchy silk of a certain species of spider behaves much like a spring, obeying Hooke's Law. Even though they are different lengths, assume each string in the figure has a force constant of 0.80 mN/cm. The spider, whose mass is 17 mg , 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

here,

mass os spider = 17 mg = 1.7*10^-5 kg
Force Constant = 0.80 mN/cm

Writing Tension in string into Components

Tbx = Tb*cos70 = Tb * 0.34 = 0.02Tb N
Tby = Tb*sin70 = Tb * 0.93 = 0.07Tb N
Tax = Ta*cos30 = Ta * 0.86 = 0.06Ta N
Tay = Ta*sin30 = Ta * 0.50 = 0.04Ta N

Tc = mg = 1.7*10^-5 *9.8
Tc = 0.000166 N = 0.16mN

Tension in X direction

0.02Tb - 0.06Ta = 0
0.06Ta = 0.02Tb

Ta = 0.333Tb ------------(1)

tension in Y direction

0.04Ta + 0.07Tb = Tcy
0.04Ta + 0.07Tb = 0.00016 ------------(2)

Using 1 in 2, we get
0.04*0.333Tb + 0.07Tb = 0.00016

Tb = 0.001920 N = 1.92 mN

so eqn 1 becomes
Ta = 0.333*0.0019

Ta = 0.00063 N = 0.63 mN

Therefore Tension in strin A,B,C is 0.63 mN,1.92 mN and 0.16 mN Respectively.

Part B:
As Force = Kx
Where
K = Force Constant
X = stretched Distance

Fa/K = Ta/K = Xa
Xa = .63 / .80
Xa = 0.87 cm

Fb/K = Tb/K = Xb
Xb = 1.92 / .80
Xb = 2.40 cm

Fc/K = Tc/K = Xc
Xc = .16 / .80
Xc= 0.20 cm

the distance each strand is stretched beyond its normal length for A,B,C is 0.87 cm, 2.40 cm , 0.20 cm respectively.

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