Please help with Physics question. I missed a lecture, and have tried to get thi

Question

Please help with Physics question. I missed a lecture, and have tried to get this correct, but keep missing it. Im guessing i have sig fis off. If you can please show all the work, and easily identifiable answers so I can follow the problem! Thanks

(a) Calculate the tension (in N) in a vertical strand of spiderweb if a spider of mass 5.00 x 10-5 kg hangs motionless on it. 45 (b) Calculate the tension (in N) in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in the figure. 5.0° 5.0° The strand sags at an angle of 12.0° below the horizontal. 8.78 Compare this with the tension in the vertical strand (find their ratio) tension in horizontal strand tension in vertical strand = 2.41

Explanation / Answer

first we always draw free body diagram
next, you have mass and need to convert it to force (use gravitational acceleration)
5.00 * 10^-5 kg (9.81m/s^2)
remember units of kg*m/s^2 = Newtons (N)

that is the only force downward and the force holding it from falling is the tension (upward force)
so sum of the forces in y = T = mg = 5*10^-5 * 9.81
T = 4.905 * 10^-4 N

part b.
draw a free body diagram again this time there are two tensions going up at twelve degree angles on either side of the spider
there is also your spider's force downward

first to solve this we need to split the force of the tension into x components and y components.
tx=Tcos(12deg)
ty = Tsin(12deg)

now we do our sum of the forces
the forces in the x direction cancel so we won't do that one
sum Fy = Tsin(12)+Tsin(12)-5*10^-5kg * 9.81 m/s^2
solve for T
T = 4.905*10^-4/(2*sin(12))
if i typed everything in right i got T=1.18*10^-3

c)

This is 2.41 times the tension in the vertical strand.

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