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Question
In Figure, the co-efficient of friction between the floor and the body B is 0.1. The co-efficient of friction between the bodies B and A is 0.2. A force F is applied as shown on B. The mass of A is m /2 and of B is m. Which of the following statements are true?
- The bodies will move together if F = 0.25 mg.
- The body A will slip with respect to B if F = 0.5 mg.
- The bodies will move together if F = 0.5 mg.
- The bodies will be at rest if F = 0.1 mg.
- The maximum value of F for which the two bodies will move together is 0.45 mg.
Solution
a, b, d and e
Explanation:
Consider the adjacent diagram. The frictional force on B(f1) and frictional force on A (f2) will be as shown.
Lat A and B are moving together `a_"common" = (F- t_1)/(m_A + m_B)`
= `(F - f_1)/((m/2) + m)`
= `(2(F - f_1))/(3 m)`
Pseudo force on `A = (m_A) xx a_"common"`
= `m_A xx (2(F - f_1))/(3m)`
= `m/2 xx (2(F - f_1))/(3m)`
= `(F - f_1))/3`
The force (F) will be maximum when
Pseudo force on A = Frictional force on A
⇒ `(F_"max" - f_1)/3 = μ m_A g`
= `0.2 xx m/2 xx g`
= 0.1 mg
⇒ `F_"max" = 0.3 mg + f_1`
= `0.3 mg + (0.1) 3/2 mg`
= 0.45 mg
⇒ Hence, the maximum force up to which bodies are together is `F_"max" = 0.45 mg`
a. Hence, for F = 0.25 mg < Fmax bodies will move together.
b. For F = 0.5 mg > Fmax, body A will slip with respect to B.
c. For F = 0.5 mg > Fmax, bodies slip.
`(f_1)_"max" = μ m_Bg = (0.1) xx 3/2 m xx g` = 0.15 mg
`(f_2)_"max" = μ m_Ag = (0.2) (m/g) g` = 0.1 mg
Hence, the minimum force required for moment of the system (A + B)
`F_"min" = (f_1)_"max" + (f_2)_"max"`
= 0.15 mg + 0.1 mg
= 0.25 mg
d. Given, force F = 0.1 mg < Fmin.
Hence, the bodies will be at rest.
e. Maximum force for combined movement Fmax = 0.45 mg.
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