(a) Calculate the tension in a vertical strand of spiderweb if a spider of mass 6.00 ✕ 10-5 kg hangs motionless on it.(b) Calculate the tension in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the tightrope walker in Figure 4.13. The strand sags at an angle of 13.0° below the horizontal.(c) Compare this with the tension in the vertical strand (find their ratio). **Title: Analyzing Forces on a Tightrope****Image Description:**In this educational illustration, we explore the forces acting on a tightrope walker. The diagram is divided into two parts, labeled (a) and (b), showing various forces and angles involved.**Diagram (a): Tightrope Walker Setup**- A person stands in the center of a horizontal tightrope.- The tightrope is slightly sagging with angles of 5° at both ends.- Forces acting on the system: - \( T_L \) and \( T_R \): Tension forces to the left and right, respectively. - \( w \): The weight force acting downward.- The coordinate axes, x (horizontal) and y (vertical), are marked.**Diagram (b): Force Balance Analysis**- A detailed vector diagram breaks down the forces: - The weight \( w \) pointing downward is balanced by the vertical components of the tension forces, \( T_{Ly} \) and \( T_{Ry} \). - Horizontal components \( T_{Lx} \) and \( T_{Rx} \) cancel out since net \( F_x = 0 \). - The inclined vectors show that the tension forces \( T_L \) and \( T_R \) are angled at 5° above the horizontal, demonstrating the equilibrium in vertical forces with net \( F_y = 0 \).**Force Components:**- \( T_L \) and \( T_R \) are decomposed into horizontal (\( T_{Lx} \), \( T_{Rx} \)) and vertical (\( T_{Ly} \), \( T_{Ry} \)) components.- Step-by-step resolution showcases vector addition ensuring vertical forces sum to zero, balancing the walker's weight.This conceptual representation aids in understanding the principles of static equilibrium and tension in physics, relevant in balancing and support structures.

Name: (a) Calculate the tension in a vertical strand of spiderweb if a spide
Uploaded: 2024-03-20T06:57:57.000Z
Channel: Bartleby
Description: (a) Calculate the tension in a vertical strand of spiderweb if a spider of mass 6.00 ✕ 10-5 kg hangs motionless on it.(b) Calculate the tension in a horizontal strand of spiderweb if the same spider sits motionless in the middle of it much like the t

When an object hangs vertically by a rope, then under the equilibrium condition, weight of the object is equal to the tension in the rope.(a) Under equilibrium condition- Tv=mgTv=6×10-5 kg×9.8 m/s2Tv=5.88×10-4 N (b) From the figure it is clear that- TRx=TRcos13 x^TLx=-TLcos13 x^ ∵∑Fx=0⇒TRx+TLx=0⇒TRcos13-TLcos13=0⇒TR=TL=Th TRy=TRsin13 y^TLy=TLsin13 y^W=-mg y^ ∵∑Fy=0⇒TRy+TLy+W=0⇒TRsin13+TLsin13-mg=0⇒2Thsin13=mg⇒Th=mg2sin13⇒Th=6×10-5kg×9.8 m/s22sin13⇒Th=13.06×10-4 N (c) Ratio of the horizontal to vertical tension is- ThTv=13.06×10-4 N5.88×10-4 NThTv=2.22 This shows that the tension in horizontal strand of spiderweb is about 2.22 times larger than the tension in vertical strand. Conclusion: (a) 5.88×10-4 N(b) 13.06×10-4 N(c) Th=2.22Tv