Find Min Force to Keep 50kg Block in Place

[rainingurl]

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.

 

[jIyajbe]

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

 

[ogb p]

The correct approach to solving this problem involves understanding the concept of equilibrium. In order for the block to remain in place, the net force acting on it must be zero. This means that the forces acting in the horizontal direction must be balanced. The minimum force required to keep the block in place is equal to the maximum static friction force that the surface can provide. This can be calculated using the equation Ff = µN, where µ is the coefficient of static friction and N is the normal force. In this case, the normal force is equal to the weight of the block, which is mg = 50kg*9.8m/s^2 = 490N. Therefore, the minimum force required to keep the block in place is Ff = µN = 0.13*490N = 63.7N. This is the minimum force that must be applied perpendicular to the wall to counteract the weight of the block and keep it in place.