Box moving up a slope by a force

  • Thread starter Thread starter BWE38
  • Start date Start date
  • Tags Tags
    Box Force Slope
AI Thread Summary
A 40.8 kg box is being pulled up a 13.5-degree slope by a 294 N force, with a coefficient of kinetic friction of 0.21. The initial calculations for acceleration yielded conflicting results, prompting a review of the free-body diagram and forces involved. The normal force was calculated as approximately 389 N, and friction was determined to be about 81.73 N. The correct approach requires accounting for the x-component of the weight using mgsinθ, which was initially overlooked. Properly applying these calculations will yield the correct acceleration of the box.
BWE38
Messages
10
Reaction score
0
A 40.8 kg box is being pulled up a 13.5 deg slope by a force of 294 N which is parallel to the slope. The coeffcient of kinetic friction between the box and the slope is 0.21. what is the acceleration of the box?
-----------------------------------------------------
Equations: F=MA, Fk=μk(MA)
-----------------------------------------------------
Attemped at the solution

deg= 13.5 M= 40.8 kg F=294 μκ= 0.21

force x-direct---------------------------------------force y-direct

294(cos13.5)(0.21)=(40.8 kg)A------------------294(sin13.5)(.21)=(40.8)A
60.03=40.8A---------------------------------------14.41=40.8A
A=1.47 m/s^2-------------------------------------A= 0.35 m/s^2

acceleration= √(1.47^2+0.35^2)
acceleration= 1.51 m/s^2
---------------------------------------------------------

The answer is 2.91 m/s^2, which I don't understand.
 
Last edited:
Physics news on Phys.org
I don't really understand your calculations here.

Have you drawn a free-body diagram with all 4 forces? Put the x-axis along the slope of the incline, and then show us what your expressions for the x and y components are.
 
I have my free body diagram attached.
 

Attachments

  • free body diagram.png
    free body diagram.png
    9.4 KB · Views: 3,225
For the y-direction ( perpendicular to slope ),
normal force = weight * cos(13.5) = 389N

For the x-direction ( up the slope ),
applied force - friction - weight_x = m*a

What do you get when you work this out?
( remember that friction = μ * normal force )
 
normal force=40.8cos13.5(9.81)=389.19N

friction=μκ(Nf)=81.73 N

x-direction= applied force - friction - weight= MA

294-81.73-40.8=40.8A
212.27 N - 40.8 kg= 40.8 kg A

am I doing everything right so far?
 
This is all good apart from the weight part - you need the x-component of the weight, ie mgsinθ.
Once you put that in, it will work out correctly.
 
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