Calculating Acceleration: Understanding Horizontal Forces on a 3kg Mass

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A 3kg mass is subjected to two horizontal forces: 9.0 N due east and 8.0 N at 62 degrees north of west. The book states that the resulting acceleration is 2.9 m/s^2, which some participants question based on their calculations. The discussion highlights the importance of considering both components of the second force, specifically the northward component, which is often overlooked. It emphasizes that both x and y components can be analyzed separately or combined using trigonometry to arrive at the correct answer. Understanding these components is crucial for accurate calculations of acceleration.
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only two horozontal forces act on a 3kg mass. one force is 9.0 N acting due east, and the other is 8.0 N acting 62 degrees north of west. what is the magintude of the body's acceleration.


The answer in the book is 2.9 m/s^2. but how can this be?

shouldn't the solution be (9 + 8cos(118))/3 = 1.74808 m/s^2 ?
 
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you are missing a component of the force

Sure you got that first part right but you're missing the fact that there is ANOTHER COMPONENT of force being appleid on this object. You decomposed that 8N force in 8cos 62 and 8sin62 what hapopened to the 8sin62?
 
What about the northward component of the second force?

The book's answer is correct.
 
damn is all i can say
 
You can deal with x and y components separately as you did, or you can use trigonometry to add the two force vectors.
 
"i can't be this dumb. can i? "

Don't ask questions like that! You might not like the answers!
 
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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