Find Min Force to Keep 50kg Block in Place

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To determine the minimum force required to keep a 50kg block of wood in place against a smooth wall with a static friction coefficient of 0.13, the frictional force must equal the weight of the block. The weight (Fg) is calculated as Fg = mg, where m is the mass and g is the acceleration due to gravity. The frictional force (Ff) can be expressed as Ff = µFn, where µ is the coefficient of static friction and Fn is the normal force. The correct approach is to set Ff equal to Fg, leading to the equation µFn = mg. This ensures that the block remains stationary against the wall.
rainingurl
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Homework Statement


a 50kg block of wood is pressed against a smooth wall by a force perpendicular to the wall. if the coefficient of static function is .13, what is the minimum force that is necessary to keep the block of wood in place.

Homework Equations


Fg= mg
Fn = ma
Ff= µ Fn = µma
Fa = ma

The Attempt at a Solution


Ff+Fn-fa = 0 since we want the block to be still.
µma +ma = fa
.13*50kg*a + 50kg*a = fa
[(.13*50)+50]a=ma
56.5a= 50a
this is wrong.
 
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Actually, what you want is for the frictional force (which points upward, in this case) to be equal in magnitude to the weight of the block of wood (which points downward). So, your equation Ff+Fn-fa = 0 is not correct; it should be just Ff - Fg = 0. You then need to use the relationship between the normal force and the frictional force.

HTH,

jIyajbe
 
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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