Is 24.01 the Correct Answer for the Static Friction Problem?

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The discussion centers on solving a static friction problem, with the answer calculated as 24.01. The user explains their reasoning, noting that since the box is stationary, the acceleration is zero, leading to the equation 0 = Fa - Ff. They calculate the frictional force using the coefficient of friction (0.245) and the normal force (98N). A question arises about the presence of two friction forces, prompting further clarification on the free body diagram used in the calculations. The conversation emphasizes understanding static friction and the importance of accurate free body diagrams in solving such problems.
Bacheeseafg
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http://imgur.com/bayjp I just need some clarification, the answer I got was 24.01, because the box is not moving, meaning a = 0, therefore 0 = Fa - Ff, I know Ff = (Mew x Fn), Mew = 0.245 and Fn = 98N, (I did 1 FBD of the sum of all masses). Meaning 0 = Fa - (0.245 x 98) and I just rearranged the question to get my answer.
 
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What did your free body diagram look like?
(Arn't there two friction forces?)
 
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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