Fnet = 45.0Find Accel. in Airport Problem with 24.0 kg & 45.0 N Force

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To find the acceleration of the 24.0 kg object being pulled at an angle with a 45.0 N force, the net force in the horizontal direction (Fx) is calculated as 41.72 N, and the vertical force (Fy) is 18.86 N. Since friction is negligible, the acceleration can be determined using Newton's second law, F=ma. The total net force acting on the object is the horizontal component, which leads to the calculation of acceleration. The discussion highlights the importance of recognizing the relationship between force and acceleration in solving the problem.
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A woman at an airport is pulling her 24.0 kg by a strap at an angle of 22 ° above the horizontal as shown in figure Fig. P5.44. She pulls on the strap with a 45.0 N force, and friction is negligible.
How would you find acceleration in this problem, there doesn't seem to be enough info but I know there is.
I calculated this
Fx = 41.72
Fy = 18.86
 
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Well... how do you relate force and acceleration...
 
lol, duh F=ma
 
thank you, I just needed that little hint
 
anytime :approve:
 
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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