Is Tension in a Tow Chain Equal to the Weight of the Car?

  • Thread starter Thread starter preluderacer
  • Start date Start date
  • Tags Tags
    Tension
AI Thread Summary
The tension in a tow chain does not equal the weight of the car when being towed at constant velocity on a horizontal road. Instead, the tension must balance the horizontal forces acting against the car's motion, as per Newton's First Law. The car's weight acts vertically downward and does not oppose the tension force. Therefore, the tension in the chain is determined by the horizontal forces, not the weight of the car. This distinction is crucial for understanding the dynamics of towing.
preluderacer
Messages
83
Reaction score
0
A car is being towed at constant velocity on a horizontal road using a horizontal chain. The tension in the chain must be equal to the weight of the car in order to maintain constant velocity.

Is this true? I have a hard time actually visualizing physics.
 
Physics news on Phys.org
Per Newton's First Law, the horizontal pulling tension must be balanced by forces acting horizontally in the opposite direction, and equal to the tension force. Does the car's weight act opposite to the tension force?
 
No, it doesnt, so i guess it isn't equal to the weight thanks!
 
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Thread 'Variable mass system : water sprayed into a moving container'
Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
Back
Top