A block of mass 1 kg is placed over another block of mass 2 kg which is placed on a horizontal smooth surface. The coefficient of friction
between the two blocks is 0.2. A force "F" is applied at an angle 30° from the horizontal (as shown in the figure below). Calculate the
maximum value of "F" such that the two blocks move together.
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Answers
Answer:
f=uR
R=mg+Fsinα
If block will slide,
then Fcosα>f
Resultant force= Fcosα−f
ma=Fcosα−u(mg+Fsinα)(m=combinedmass)
ma=Fcosα=u(mg+Fsinα)
ma=
2
F
3
−
2
1
(3×10+
2
F
)
3a=
2
F
3
−
2
1
(
2
60+F
)
3a=
2
F
3
−
4
60+F
3a=
4
2F
3
−60−F
⇒a=
12
F(2−
3
−1)−60
m/s
Answer: Maximum value of F, such that the two blocks move together is 4√3/(1+0.2√3) N
Explanation:
From the free body diagram we can write the following equations:
√3/2 F - f = 2.a -------(1)
Assuming no slipping between the blocks.
f = 1.a --------(2)
from (1) and (2) we get
√3/2 F - a = 2a
=> F = 2√3 a N -------(3)
f = u . mg (where u = friction coefficient. )
=> f = 0.2 x (1 x 10 - F/2) = 2 - 0.1F N
from eq (2)
f = a
=> a = 2 - 0.1Fm/s²
From eq (3)
F = 2√3 x(2 - 0.1F) N
=> F = 4√3/(1+0.2√3) N
Therefore, the maximum value of F, such that the two blocks move together is 4√3/(1+0.2√3) N
