In the figure masses m1, m2 and M are 20 kg, 5 kg and 50 kg respectively. The coefficient of frictio

Q: Wait, why does the string exert zero net force on block M?

Look at the pulleys attached to M. At the top pulley, the string pulls M to the right and down. At the bottom pulley, the string pulls M to the left and up. The rightward pull perfectly cancels the leftward pull, and the downward pull perfectly cancels the upward pull! The string is essentially a ghost to block M.

Q: Why did we assume $m_2$ is experiencing maximum kinetic friction?

We proved that the system cannot be at rest (because if $a=0$, $f_1$ must equal $f_2$, which contradicts the given $f_1 = 2f_2$). Since the system is moving, $m_2$ is sliding across the ground, meaning it must experience its maximum kinetic friction.

Q: How do we know $m_1$ doesn't slip on M?

We calculated the required friction $f_1$ to be 30 N (since $f_1 = 2f_2$ and $f_2 = 15$ N). Then, we calculated the maximum possible static friction for $m_1$, which is $\mu m_1 g = 60$ N. Since the required 30 N is well below the 60 N limit, the static friction is strong enough to prevent slipping.

Q: Why is the friction $f_1$ on $m_1$ directed to the right?

Block M is being pulled to the right by force F. Because of inertia, $m_1$ wants to stay behind, meaning it tends to slide to the left relative to M. Friction always opposes relative motion, so the friction $f_1$ on $m_1$ acts to the right.

Q: Why are the accelerations of $m_1$ and $m_2$ equal?

If you write the length of the string as $L = (x_1 - x_M) + h + (x_M - x_2)$, the $x_M$ terms cancel out! This leaves $L = x_1 - x_2 + h$. Since the length is constant, the distance between $m_1$ and $m_2$ is constant, meaning their velocities and accelerations must be identical.

Question

In the figure masses $m_1$, $m_2$ and $M$ are $20	ext{ kg}$, $5	ext{ kg}$ and $50	ext{ kg}$ respectively. The coefficient of friction between $M$ and ground is zero. The coefficient of friction between $m_1$ and $M$ and that between $m_2$ and ground is $0.3$. The pulleys and the strings are massless. The string is perfectly horizontal between $P_1$ and $m_1$ and also between $P_2$ and $m_2$.
The string is perfectly vertical between $P_1$ and $P_2$. An external horizontal force $F$ is applied to the mass $M$.
(Take $g = 10	ext{ m/s}^2$.)
(a) Draw a free body diagram of mass $M$, clearly showing all the forces.
(b) Let the magnitude of the force of friction between $m_1$ and $M$ be $f_1$ and that between $m_2$ and ground be $f_2$. For a particular force $F$ it is found that $f_1 = 2f_2$. Find $f_1$ and $f_2$. Write equations of motion of all the masses. Find $F$, tension in the string and accelerations of the masses.

Sigma Percentile · Accepted Answer

### The Illusion of Complexity

At first glance, this problem looks like an absolute nightmare. We have three different blocks, two pulleys, multiple rough surfaces, and an external force pulling the entire system. It is easy to get lost in a sea of equations. But physics is rarely about brute force algebra; it is about finding the hidden symmetries and constraints that make the complexity collapse into simplicity. Let's break this down step by step.

### The Kinematic Secret

The most powerful tool in our arsenal is the string constraint. Let's write down the total length of the string $L$ in terms of the horizontal coordinates of the blocks. Let $x_1$ be the position of $m_1$, $x_2$ be the position of $m_2$, and $x_M$ be the position of the large block $M$.

The string goes from $m_1$ to the first pulley $P_1$ (a distance of $x_1 - x_M$), then vertically down to $P_2$ (a constant distance $h$), and finally from $P_2$ to $m_2$ (a distance of $x_M - x_2$). Adding these up, we get:

$$L = (x_1 - x_M) + h + (x_M - x_2)$$

Notice something magical? The $x_M$ terms perfectly cancel each other out! The length of the string simplifies to $L = x_1 - x_2 + h$. Because the length of the string is constant, the distance between $m_1$ and $m_2$ must remain constant. If we differentiate this twice with respect to time, we arrive at a beautiful conclusion: $a_1 = a_2$. The accelerations of $m_1$ and $m_2$ relative to the ground are always identical, regardless of what the large block $M$ is doing!

### The Pulley Paradox

Now, let's look at the forces acting on the large block $M$. It is being pulled to the right by the external force $F$, and it experiences friction $f_1$ from $m_1$. But what about the pulleys?

At the top pulley $P_1$, the string pulls $M$ to the right and downwards with tension $T$. At the bottom pulley $P_2$, the string pulls $M$ to the left and upwards with tension $T$. If we sum the horizontal forces exerted by the string on $M$, we get $T$ (right) $- T$ (left) $= 0$. The vertical forces also cancel out. The string is essentially a ghost to block $M$; it exerts **zero net force** on it!

### The Friction Limits and Proof of Motion

Before we write our dynamic equations, we must establish our boundaries. The maximum static friction for $m_1$ is $(f_1)_{	ext{max}} = \mu m_1 g = 0.3 	imes 20 	imes 10 = 60	ext{ N}$. For $m_2$, it is $(f_2)_{	ext{max}} = \mu m_2 g = 0.3 	imes 5 	imes 10 = 15	ext{ N}$.

Could the blocks be stationary? If they were at rest, their accelerations would be zero. The tension $T$ would have to balance the friction on both blocks, meaning $T = f_1$ and $T = f_2$. This implies $f_1 = f_2$. However, the problem explicitly states that $f_1 = 2f_2$. This is a glaring contradiction! Therefore, the blocks cannot be at rest; they must be accelerating.

### The No-Slip Revelation

Since the system is moving, $m_2$ is sliding across the ground. This means it experiences its maximum kinetic friction, so $f_2 = 15	ext{ N}$. Using the given condition $f_1 = 2f_2$, we immediately find that $f_1 = 30	ext{ N}$.

Now we must ask a critical question: is $m_1$ slipping on $M$? We just found that the required friction $f_1$ is $30	ext{ N}$. But we calculated earlier that the maximum static friction available is $60	ext{ N}$. Since $30	ext{ N}$ is well below the $60	ext{ N}$ limit, $m_1$ does **not** slip on $M$. They move together as a single unit!

This means $a_1 = a_M$. And since we already proved $a_1 = a_2$, we can conclude that all three masses share the exact same acceleration $a$.

### The Final Execution

We are now ready to write Newton's second law for $m_1$ and $m_2$. For $m_1$, the friction $f_1$ pulls it to the right, and tension $T$ pulls it to the left:

$$30 - T = 20a$$

For $m_2$, tension $T$ pulls it to the right, and friction $f_2$ pulls it to the left:

$$T - 15 = 5a$$

Adding these two equations eliminates the tension $T$, leaving us with $15 = 25a$, which gives an acceleration of $a = 0.6	ext{ m/s}^2$. Substituting this back into either equation yields a tension of $T = 18	ext{ N}$.

Finally, we write the equation of motion for the large block $M$. The only horizontal forces acting on it are the external force $F$ to the right and the friction $f_1$ to the left (remember, the string forces canceled out!).

$$F - f_1 = M a$$

Substituting our known values:

$$F - 30 = 50(0.6)$$

$$F - 30 = 30$$

$$F = 60	ext{ N}$$

And just like that, the illusion of complexity shatters, leaving behind a beautifully elegant solution.

Visualized Solution

The System Architecture

String Constraint \& Accelerations

Net String Force on $M$

Limiting Friction Values

Can the Blocks be Stationary?

Kinetic Friction on $m_{2}$

Checking Relative Motion

Equations of Motion

Solving for $a$ and $T$

Equation of Motion for $M$

The Sigma Insight: Static and Kinetic Friction

The Illusion of Complexity

The Kinematic Secret

The Pulley Paradox

The Friction Limits and Proof of Motion

The No-Slip Revelation

The Final Execution

Similar Questions

Two masses $m_{1} = 5 kg$ and $m_{2} = 10 kg$ connected by an inextensible string over a frictionless pulley, are moving as shown in the figure. The coefficient of friction of horizontal surface is $0.15$ . The minimum weight $m$ that should be put on top of $m_{2}$ to stop the motion is

List-I

List-II