A sledge loaded with bricks has a total mass of 18.0 kg

  • Thread starter Thread starter imaqueen
  • Start date Start date
  • Tags Tags
    Mass
AI Thread Summary
To determine the tension in the rope pulling a sledge loaded with bricks, the equation T = mg(sinθ) is used, where T is tension, m is mass, g is gravitational acceleration (9.8 m/s²), and θ is the angle of inclination (20 degrees). In this scenario, with a total mass of 18.0 kg and assuming no friction, the tension calculates to approximately 30.9 N. The tension is influenced by the mass of the sledge and the angle of the rope; higher mass and angle increase tension. If friction were present, the equation would need to account for the coefficient of friction as well. Understanding these principles is crucial for solving similar physics problems effectively.
imaqueen
[?] A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at a constant speed by a rope. the rope is inclined at 20 deg above the horizon surface and the sledge moves a distance of 20.o m on a horizontal surface. what is the tension of the rope. How do you find tension is there an equation for it or does it just depend on the problem. help would be greatly appreciated
 
Last edited by a moderator:
Physics news on Phys.org
The tension in the rope is the force the rope exerts on the sledge.

You do have one problem: no matter how heavy the sledge is, if there is NO friction, it is not necessary to exert ANY force to keep the sledge moving. Are you given a "coefficient of friction"? If so then the horizontal force will be that coefficient times the WEIGHT of the sledge (not its mass). After you have found that, use trigonometry to find the force the rope has to exert. (The rope is the hypotenuse of a right triangle. The horizontal force is the horizontal leg of that triangle.)
 


To find the tension of the rope, we can use the equation T = mg(sinθ + μcosθ), where T is the tension, m is the mass of the sledge, g is the acceleration due to gravity (9.8 m/s^2), θ is the angle of inclination, and μ is the coefficient of friction. In this case, μ can be assumed to be 0 since the sledge is moving at a constant speed. Therefore, the equation simplifies to T = mg(sinθ). Plugging in the values, we get T = (18.0 kg)(9.8 m/s^2)(sin20°) = 31.3 N. This is the tension in the rope that is pulling the sledge at a constant speed on the horizontal surface.
 


To find the tension in the rope, we can use the equation T = mgsinθ, where T is the tension, m is the mass of the sledge, g is the acceleration due to gravity (9.8 m/s^2), and θ is the angle of inclination of the rope.

In this case, T = (18.0 kg)(9.8 m/s^2)sin 20° = 30.9 N.

The tension in the rope is dependent on the mass of the sledge, the angle of inclination, and the acceleration due to gravity. In general, the greater the mass and angle of inclination, the greater the tension will be. However, if the sledge is not moving or is moving at a constant speed, the tension in the rope will be equal to the force of friction acting on the sledge.

I hope this helps to clarify how to find tension in a problem. If you have any further questions, please don't hesitate to ask.
 
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}...
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...
Back
Top