Masses M1, M2 and M3 are connected by strings of negligible mass which passes over massless and fric

Q: Wait, why is the acceleration zero if the blocks are clearly moving?

Uniform velocity means the speed and direction are constant. According to Newton's First Law, a constant velocity implies zero net force, which means zero acceleration. Motion doesn't require a net force; only a *change* in motion does!

Q: Why don't we include the tensions when writing the equation for the whole system?

Tensions are internal forces when you consider all three masses as a single system. For every tension pulling one way on one block, it pulls the exact opposite way on the adjacent block, completely canceling out in the macroscopic view.

Q: How do we know the exact direction of friction on $M_2$ and $M_3$?

Friction always opposes relative sliding. Since $M_1$ moves downwards, it pulls $M_2$ up the incline and $M_3$ to the left. Therefore, friction on $M_2$ acts down the incline, and friction on $M_3$ acts to the right.

Q: Why is the normal force on $M_2$ equal to $M_2 g \cos 37^\circ$ instead of just $M_2 g$?

On an inclined plane, gravity has two components: one parallel to the plane ($mg \sin\theta$) and one perpendicular to it ($mg \cos\theta$). Since the block doesn't accelerate perpendicular to the plane, the normal force must perfectly balance the perpendicular component of gravity.

Q: Can we solve this by writing individual equations for each block instead of a system equation?

Absolutely! You can write $M_1 g - T_1 = 0$, $T_1 - T_2 - M_2 g \sin 37^\circ - f_{k2} = 0$, and $T_2 - f_{k3} = 0$. Adding these three equations together will give you the exact same master equation we used, proving that both methods are perfectly valid.

Question

Masses $M_1, M_2$ and $M_3$ are connected by strings of negligible mass which passes over massless and frictionless pulleys $P_1$ and $P_2$ as shown in figure. The masses move such that the portion of the string between $P_1$ and $P_2$ is parallel to the inclined plane and the portion of the string between $P_2$ and $M_3$ is horizontal. The masses $M_2$ and $M_3$ are $4.0	ext{ kg}$ each and the coefficient of kinetic friction between the masses and the surfaces is $0.25$. The inclined plane makes an angle of $37^\circ$ with the horizontal. If the mass $M_1$ moves downwards with a uniform velocity, find (a) the mass of $M_1$, (b) the tension in the horizontal portion of the string. (Take $g = 9.8	ext{ m/s}^2$, $\sin 37^\circ \cong 3/5$)

Sigma Percentile · Accepted Answer

### The Illusion of Motion

When we see a complex system of pulleys and masses in motion, our instinct is to immediately think about acceleration. However, the golden phrase in this problem is **uniform velocity**.

This changes everything. According to Newton's First Law, a constant velocity means the net acceleration of the system is exactly zero. Consequently, the net force acting on the entire system must also be zero. This profound realization transforms a dynamic problem into a beautiful exercise in static equilibrium.

### Breaking Down the System: The Horizontal Mass

Let's begin by isolating the mass $M_3$ resting on the horizontal surface. Since the hanging mass $M_1$ is moving downwards, it pulls the entire string, causing $M_3$ to slide to the left.

To maintain a constant velocity, the forces acting horizontally on $M_3$ must perfectly cancel each other out. The tension $T_2$ pulls it to the left, while the kinetic friction $f_{k3}$ opposes this motion to the right.

$$T_2 = f_{k3} = \mu M_3 g$$

By substituting the given values, we can easily find the tension in this horizontal segment of the string:

$$T_2 = (0.25)(4.0)(9.8) = 9.8 	ext{ N}$$

### The Inclined Mass: Fighting Gravity and Friction

Next, we shift our focus to the mass $M_2$ situated on the inclined plane. This block is being dragged up the $37^\circ$ slope.

It faces resistance from two distinct forces. First, gravity pulls it down the incline with a component equal to $M_2 g \sin 37^\circ$. Second, the rough surface exerts a kinetic friction $f_{k2}$ that also acts downwards along the plane.

To calculate this friction, we must first determine the normal force. On an inclined plane, the normal force balances the perpendicular component of gravity:

$$N_2 = M_2 g \cos 37^\circ$$

Therefore, the kinetic friction opposing the upward motion is:

$$f_{k2} = \mu M_2 g \cos 37^\circ$$

### The Master Equation: Tying It All Together

While we could write individual equations for each mass and solve for the tensions, there is a much more elegant approach. We can treat all three masses as a single, unified system.

In this macroscopic view, the internal tensions ($T_1$ and $T_2$) perfectly cancel each other out. We only need to balance the external driving force against the external opposing forces. The sole driving force propelling the system forward is the weight of the hanging mass, $M_1 g$.

This driving force must overcome the combined resistance of the other two masses:

$$M_1 g = M_2 g \sin 37^\circ + f_{k2} + f_{k3}$$

Substituting our expressions for friction, we arrive at the master equation:

$$M_1 g = M_2 g \sin 37^\circ + \mu M_2 g \cos 37^\circ + \mu M_3 g$$

### The Final Calculation

Notice how the acceleration due to gravity, $g$, appears in every single term. We can elegantly divide the entire equation by $g$, simplifying our calculation significantly:

$$M_1 = M_2 \sin 37^\circ + \mu M_2 \cos 37^\circ + \mu M_3$$

Now, we carefully substitute the numerical values provided in the problem. We know that $\sin 37^\circ \approx \frac{3}{5}$ and $\cos 37^\circ \approx \frac{4}{5}$:

$$M_1 = 4\left(\frac{3}{5}ight) + 0.25(4)\left(\frac{4}{5}ight) + 0.25(4)$$

Breaking down the arithmetic step-by-step:

$$M_1 = 2.4 + 0.8 + 1.0$$

Adding these values together yields the final mass required to keep the system moving at a uniform velocity:

$$M_1 = 4.2 	ext{ kg}$$

The beauty of physics lies in how complex, interconnected systems can be unraveled using fundamental conservation principles and a bit of algebraic elegance.

Visualized Solution

$S y s t e m O v er v i e w$

$C o n d i t i o n f or U ni f or m V e l oc i t y$

$F or ces o n M_{3}$

$T e n s i o n in H or i z o n t a l S t r in g$

$V a l u e o f T_{2}$

$F or ces o n M_{2}$

$F r i c t i o n o n I n c l in e$

$F or ces o n M_{1}$

$N e t F or ce o n S y s t e m$

$E q u a t in g F or ces$

$S u b s t i t u t in g V a l u es$

$C a l c u l a t in g M_{1}$

The Sigma Insight: Static and Kinetic Friction

The Illusion of Motion

Breaking Down the System

The Inclined Mass

The Master Equation

The Final Calculation

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

A block rests on a rough inclined plane making an angle of $3 0^{\circ}$ with the horizontal. The coefficient of static friction between the block and the plane is $0.8$ . If the frictional force on the block is $10 N$ , the mass of the block (in $kg$ ) is ( $g = 10 m/s^{2}$ )

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } System Overview

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Condition for Uniform Velocity

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces on M3​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Tension in Horizontal String

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Value of T2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces on M2​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Friction on Incline

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Forces on M1​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Net Force on System

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Equating Forces

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Substituting Values

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Calculating M1​