Solve Tension Problem: A Sledge w/ Bricks & Coefficient of Kinetic Friction

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To solve the tension problem involving a sledge loaded with bricks, it's essential to understand that the sledge moves at constant speed, indicating zero acceleration. This means the total forces acting on the sledge must balance out to zero. Participants suggest drawing a diagram to visualize the forces, including tension, weight, and friction, to better understand how they interact. The coefficient of kinetic friction and the angle of the rope are critical in calculating the tension. Overall, a systematic approach to diagramming forces will clarify the solution process.
mrrocketknigh
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Here's the problem: A sledge loaded with bricks has a total mass of 18.0 kg and is pulled at constant speed by a rope inclined at 20.0(degrees) above the horizontal. The sledge moves a distance of 20.0m on a horizontal suface. The coefficient of kinetic friction between the sledge and surface is 0.500. (a) What is the tension of the rope?


I don't know how to even start this problem, or how to find out what the tension is. Can anybody help me?
 
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welcome to pf!

hi mrrocketknigh! welcome to pf! :wink:

if the speed is constant, then the acceleration is zero, so Ftotal = ma tells us that Ftotal = 0

so draw all the forces on the sledge, and get them to add to zero …

what do you get? :smile:​
 
What do you mean by "draw all the forces on the sledge and get them to add to zero"? I'm lost.
 
Last edited:


Wait... is it 0? I tried doing it myself this whole time, but to no avial...
 
mrrocketknigh said:
What do you mean by "draw all the forces on the sledge and get them to add to zero"? I'm lost.

draw a diagram of the sledge, then draw all the forces on it …

what are they?​
 
Thread 'Variable mass system : water sprayed into a moving container'

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}...
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