Two blocks of masses M1 and M2 are connected with a string passing over a pulley as shown in figure. The block M1 lies on a horizontal surface. The coefficient of friction between the block M1 and the horizontal surface is μ The system accelerates. What additional mass m should be placed on the block M1 so that the system does not accelerate?

Question

Two blocks of masses M1 and M2 are connected with a string passing over a pulley as shown in figure. The block M1 lies on a horizontal surface. The coefficient of friction between the block M1 and the horizontal surface is μ The system accelerates. What additional mass m should be placed on the block M1 so that the system does not accelerate? <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/physics/21436f27df9a611231a41c3fe2be513d-.png" /> (A) (M2-M1/μ) (B) (M2/μ)-M1 (C) M2-(M1/μ) (D) (M2-M1)μ

Tardigrade Admin · Accepted Answer

For the equilibrium of block of mass M1: Frictional force, f = tension in the string, T where T=f =μ (m+M1)gt .....(i) For the equilibrium of block of mass M2: T=M2g .....(ii) From (i) and (ii) μ(m+M1)g=M3g m=(M2/μ)-M1

$\frac{M _{2} - M _{1}}{μ}$

$\frac{M _{2}}{μ} - M_{1}$

$M_{2} - \frac{M _{1}}{μ}$

$(M_{2} - M_{1}) μ$

For the equilibrium of block of mass $M_{1} :$

Frictional force, f = tension in the string, T

where $T = f = μ (m + M_{1}) g t ..... (i)$

For the equilibrium of block of mass $M_{2} :$

$T = M_{2} g .....$ (ii)

From (i) and (ii) $μ (m + M_{1}) g = M_{3} g$

$m = \frac{M _{2}}{μ} - M_{1}$

μM2​−M1​​

μM2​​−M1​

M2​−μM1​​

(M2​−M1​)μ

For the equilibrium of block of mass M1​: Frictional force, f = tension in the string, T where T=f=μ(m+M1​)gt.....(i) For the equilibrium of block of mass M2​: T=M2​g.....(ii) From (i) and (ii) μ(m+M1​)g=M3​g m=μM2​​−M1​

$\frac{M _{2} - M _{1}}{μ}$

$\frac{M _{2}}{μ} - M_{1}$

$M_{2} - \frac{M _{1}}{μ}$

$(M_{2} - M_{1}) μ$

For the equilibrium of block of mass $M_{1} :$

Frictional force, f = tension in the string, T

where $T = f = μ (m + M_{1}) g t ..... (i)$

For the equilibrium of block of mass $M_{2} :$

$T = M_{2} g .....$ (ii)

From (i) and (ii) $μ (m + M_{1}) g = M_{3} g$

$m = \frac{M _{2}}{μ} - M_{1}$