Find acceleration when you only have a Fx

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To find acceleration when only given a horizontal force (Fx) at a 25-degree angle, it is essential to clarify that the force is acting on an object, not being pulled itself. The force can be broken down into horizontal (Fx) and vertical (Fy) components, with Fy calculated as Fy = Fx(tan(25)). To determine acceleration, the mass of the object must also be known. The relationship between force, mass, and acceleration is governed by Newton's second law, which states that acceleration equals force divided by mass. Understanding these components is crucial for accurately calculating acceleration.
monkeypretzels
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how could you possibly find accerlation when you only have a Fx and that force is being pulled at 25 degrees :eek:
 
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First of all, a "force" cannot be pulled! I presume you mean that an object is being pulled, at 25 degrees by a force with horizontal component Fx. 25 degrees to what? The x-axis? Assuming, since Fx= Fcos(25) where F is the magnitude of the force, F= Fx/cos(25) so Fy, the vertical component, is
Fy= Fsin(25)= Fx(sin(25)/cos(25))= Fx tan(25). In order to find the acceleration, you will also need to know the mass of the object.
 
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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