The two blocks in the figure, of mass m and M respectively, are not attached, but there is a coefficient of static friction u between them. The surface below the heavier block M is frictionless. What is the minimum force magnitude F required to keep the smaller block m from slipping down? Express your answer in terms of m, M, u, and g.
Figure:
______
___| |
F-->| m | M |
|___| |
|______|
Figure:
---------______
-----___|------|
F-->|-m-|--M---|
----|___|------|
--------|______|
That's still not right but hopefully you can understand it. I wish it would just let me post a link to an image.
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on the whole system:
F = (m+M)a
on block m, call Fb the force between blocks:
F - Fb = m a
so
(m+M)a - Fb = m a
so
Fb = M a
for no slip:
Fb u = m g
Fb = m g/u
M a = m g/u
so
a = (m/M) (g/u)
so using F = (m+M)a
F = (m+M)(m/M)(g/u)
To find the minimum force magnitude (F) required to keep the smaller block (m) from slipping down, we need to consider the forces acting on both blocks.
First, let's identify the forces acting on block m:
1. Weight (mg): This force acts vertically downward and has a magnitude equal to the mass (m) of block m multiplied by the acceleration due to gravity (g).
2. Frictional force (f): The block m will experience a frictional force opposing its tendency to slide down the incline. The magnitude of the frictional force can be determined using the coefficient of static friction (u) and the normal force (N) acting on the block m.
Next, let's consider the forces acting on block M:
1. Weight (Mg): This force acts vertically downward and has a magnitude equal to the mass (M) of block M multiplied by the acceleration due to gravity (g).
2. Force applied by block m (F): This force is applied horizontally by block m and acts on block M.
Since the surface below block M is frictionless, there is no frictional force acting on block M.
To prevent block m from slipping down, the frictional force (f) acting on it must be equal to or greater than the force component (F_x) of the applied force (F) acting in the horizontal direction.
Now, let's calculate the frictional force (f):
The normal force (N) acting on block m is equal to the weight of block m minus the weight of block M.
N = mg - Mg
The maximum static frictional force (f_max) can be found by multiplying the coefficient of static friction (u) by the normal force (N).
f_max = u * (mg - Mg)
Since the block m is at the verge of slipping, the frictional force (f) will be equal to the maximum static frictional force (f_max):
f = f_max = u * (mg - Mg)
To prevent block m from slipping down, the frictional force (f) must be equal to or greater than the force component (F_x) of the applied force (F), so we have:
F_x ≤ f
F_x ≤ u * (mg - Mg)
Therefore, the minimum force magnitude (F) required to keep the smaller block (m) from slipping down is:
F = u * (mg - Mg)
We have expressed the answer in terms of m, M, u, and g.