How to Find the Friction Force on a Wedge

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To find the friction force on a wedge, the total normal force is calculated as 263.508 Newtons to the right, derived from a combination of a 210N force and the weight component of a 21kg mass. The friction force on the left side of the wedge is determined to be 79.05N by multiplying the normal force by the coefficient of friction (0.3). The normal force on the right side is calculated to be 267.57N, with specific x and y components identified. To solve for the resultant force, the y components of the frictional force and the normal force on the right side must be combined. Understanding the forces acting on both sides of the wedge is crucial for accurately determining the friction force.
SteelDirigibl
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Homework Statement


Screenshot2011-04-19at104514PM.png



Homework Equations


F=mu*N


The Attempt at a Solution



The total normal force on the wedge is going to be the 210N plus 21kg*9.8*0.26 which is 263.508 Newtons to the right. Multiply that by 0.3 to get79.05N for the friction force up on the left side of the wedge.the normal force on the right side comes to 267.57 N, which has x component 263.508 and y component 45.8

For P you add the frictional force's y components plus the y component of the normal force on the right side? I can't seem to get the answer (I did by guessing, its 212N) but I don't know how to get there.
 
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Hi SteelDirigibl! :smile:

(have a mu: µ :wink:)
SteelDirigibl said:
The total normal force on the wedge is going to be the 210N plus 21kg*9.8*0.26 which is 263.508 Newtons to the right*…

(I haven't tried this myself :redface:, but anyway …)

Have you taken into account the horizontal force on the right side of the wedge?

I think you need to start on that side, and work your way across …

the acceleration perpendicular to that slope is zero, so try starting with components along the slope. :wink:
 
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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