Tension force of an object on a slope

AI Thread Summary
The discussion revolves around calculating the tension force in a rope pulling a box up a 30-degree slope, with a coefficient of friction of 0.25 and an acceleration of 1.5 m/s² down the hill. Initial calculations for gravitational force, normal force, and frictional force were presented, but attempts to apply these in the net force equations yielded incorrect results. Participants suggest breaking the forces into components parallel and normal to the slope for clarity. The correct approach involves redefining coordinates to simplify the equations. The expected answer for the tension force is approximately 2.5 N, which has not yet been achieved by the original poster.
Dakkers
Messages
17
Reaction score
0

Homework Statement


A box is trying to be pulled by a rope up a hill that is 30 degrees above the horizontal. The coefficient of friction (\mu) is 0.25. The acceleration is 1.5m/s^{2} down the hill. The mass of the box is 2.0 kg. Find the force of tension in the rope.

Homework Equations


Fg = mg
Fn = Fg
Ff = \muFn
Fnet = ma

The Attempt at a Solution


I started by calculating each of the forces that I could.

Fg = mg = (2.0)(9.8) = 19.6 N

Fn = Fg = 19.6 N

Ff = \muFn = (0.25)(19.6) = 4.9 N

I drew a diagram and broke everything into components.

[URL]http://img231.imageshack.us/i/physicsforces.png/[/URL]

I forgot to add that if we do components of each force we have to do the components of acceleration, which becomes either 1.5cos30
or 1.5sin30 [down].

I plugged the components into the Fnet = ma equation but it didn't work, like so:

Fnet = ma (up/down)
Ftsin30 + 19.6sin60 - 19.6 - 4.9sin30 = (2.0)(-1.5sin30)
Ftsin30 - 5.0759... = -1.5
Ftsin30 = 3.5759
Ft = 7.1518

buuut that's not right because, at the least, Ff > Ft since the box is sliding down the hill.

Fnet = ma (left/right)
Ftcos30 - 4.9cos30 - 19.6cos60 - 0 = (2.0)(1.5cos30)
Ftcos 30 + 5.5564... = 2.5980...
Ftcos 30 = -2.95...
Ft = -3.41

once again, wrong.

pleeeease help!​
 
Last edited by a moderator:
Physics news on Phys.org
I haven't checked your calculations, but this question would be much easier if you broke the force up into a component parallel to the slope, and a component normal to the slope. That way, you'll get two nice equations--one for each component--that are easy to solve.
 
not to sound like a complete noob buuut what do you mean by parallel to the slope and normal to the slope?
 
You can redefine your coordinates so that the x coordinate represents distance down the slope, and the y coordinate represents distance perpendicular to the slope. I've attached a drawing to try to explain.
 

Attachments

  • bad_drawing.png
    bad_drawing.png
    851 bytes · Views: 551
yeah and I made each force into its respective components (as the ugly math says up there) and tried isolating Ft but it was wrong every time. the answer should be about 2.5 (apparently) and I haven't gotten that yet

f
 
What do you mean by isolate. There will always be other forces involved.
What are the forces acting parallel to the surface of incline?
 
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