Horizontal Force Needed to Slide 160N Crate with Friction Coefficient of .25

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To determine the horizontal force needed to slide a 160N crate with a friction coefficient of 0.25, the frictional force must first be calculated using the equation f = u * n, where n is the normal force. The normal force for a horizontal surface is equal to the weight of the crate, which is 160N. Thus, the frictional force is 0.25 * 160N, equaling 40N. This means a horizontal force of at least 40N is required to overcome friction and move the crate. A free body diagram can help visualize the forces acting on the crate for a clearer understanding of the problem.
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Homework Statement


if the coefficient of the friction between a crate and the floor is .25 and a 160 nt crate must be moved, the horizontal force need to slide the crate is?


Homework Equations



f=ma
u=f/n

The Attempt at a Solution



no idea how to begin
 
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wwangsta said:

Homework Statement


if the coefficient of the friction between a crate and the floor is .25 and a 160 nt crate must be moved, the horizontal force need to slide the crate is?


Homework Equations



f=ma
u=f/n

The Attempt at a Solution



no idea how to begin


You'll need to show some idea of what's going on before you can receive help.

How does the given information fit into the equations you gave? Have you drawn a free body diagram of the forces on the crate?
 
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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