What limit must the force excede if we want to separate two bodies + friction

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SUMMARY

The minimum force required to separate a weight from a plank on a leveled surface is 14.7 N. This calculation is based on the mass of the plank (M = 2 kg) and the weight (m = 1 kg), with friction coefficients of k1 = 0.4 between the plank and the base, and k2 = 0.3 between the weight and the plank. The total force is derived from the sum of the forces needed to overcome both frictional forces acting on the plank and the weight. Therefore, a force exceeding 14.7 N is necessary for the weight to slip off the plank.

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  • Familiarity with the concept of normal force (N = mg)
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Homework Statement



On a leveled base is a plank with a mass M = 2 kg, on the plank is a weight with a mass m = 1 kg. What limit must the force exceed, when we pull the plank in a horizontal direction for the weight to slip from the plank? Ratio of friction between the plank and the base is k1 = 0.4 and ratio of friction between the weight and the plank is k2 = 0.3.


Homework Equations



N= mg
Fk= μkN

The Attempt at a Solution



M= 2 kg
m= 1 kg
μk1= 0.4
μk2= 0.3

For the plank and weight to start moving: F1= (M+m)gμk1= 11.76 N
For the weight to start moving on the plank: F2= mgμk2= 2.94 N

So if my logic works correctly I simply add the two forces and the force will be big enough for the weigh to start falling off the plank.

F= F1 + F2= 11.76N + 2.94 N= 14.7 N

And the answer would be: The force have to be bigger than 14.7 N for the weight to fall off the plank.
 
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