So i did this question and got an answer of 43 N i was wondering if

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The discussion revolves around calculating the smallest force needed to keep a 22 kg box at rest on a 45-degree ramp, given specific coefficients of friction. The initial answer provided is 43 N, prompting others to verify the calculation. Participants confirm the answer and express agreement with the methodology used, which involves breaking down the forces into x and y components. The conversation emphasizes the importance of checking calculations and understanding the physics behind the problem. Overall, the community collaborates to ensure the accuracy of the solution.
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So i did this question and got an answer of 43 N i was wondering if someone else could try it and tell me what they get.. it its not the same ill explain my steps and see where i went wrong.

Homework Statement


A box with a mass of 22 kg is at rest on a ramp inclined at 45 to the horizontal. The coefficients of friction between the box and the ramp are Us = 0.78 and Uk = 0.65.
Determine the magnitude of the smallest force that can be applied onto the top of the box, perpendicular to the ramp, if the box is to remain at rest.


Homework Equations


Fnet = 0
Ff = UFn

The Attempt at a Solution


I broke the question down into its x and y components.
 
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Yes, that's what I get as well.
 
okay good! thank you!
 
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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