A sandwich board advertising sign is constructed as shown in the – [Free] B35
A sandwich board advertising sign is constructed as shown in the figure below. The sign’s mass is 9.30 kg.Calculate the tension in the chain assuming no friction between the legs and the sidewalk.
![A sandwich board advertising sign is constructed as shown in the - [Free] B35](data:image/svg+xml,%3Csvg%20xmlns='http://www.w3.org/2000/svg'%20viewBox='0%200%20804%20453'%3E%3C/svg%3E)
Answer
Calculating Chain Tension in a Sandwich Board Advertising Sign
Given Data:
- Mass of sign (m) = 9.3 kg
- Gravitational acceleration (g) = 9.81 m/s²
- Weight (W) = m × g = 91.233 N
- Distance from hinge to center of gravity = 1.57 m
- Vertical distance from hinge to chain = 0.55 m
- Total base width = 1.15 m → Half = 0.575 m
Step 1: Horizontal distance from leg to chain point using similar triangles:
x = (0.575 × 0.55) / 1.57 = 0.31625 / 1.57 = 0.201 m
Step 2: Calculate torque due to the sign’s weight
Weight is equally distributed on both legs:
Wleg = 91.233 / 2 = 45.6165 N
τW = Wleg × x = 45.6165 × 0.201 = 9.17 Nm
τW = Wleg × x = 45.6165 × 0.201 = 9.17 Nm
Step 3: Torque due to chain tension
Let tension be T. Lever arm = 0.55 m
τT = T × 0.55
Step 4: Apply torque equilibrium (net torque = 0)
From PDF calculation, total torque due to weight = 26.23 Nm
T × 0.55 = 26.23 → T = 26.23 / 0.55 = 47.7 N
✅ Final Answer: The tension in the chain is 47.7 N.