A string of negligible mass going over a clamped pulley of mass m supports a block of mass M as show

Q: Wait, why do we consider the tension acting twice on the pulley?

Great question! The string wraps around the pulley. It pulls the pulley downwards on the side where the block hangs, and it pulls the pulley horizontally on the side attached to the wall. Both of these are distinct, real forces exerted by the string on the pulley's surface.

Q: Why doesn't the pulley rotate if there are two tension forces pulling on it?

The two tension forces act tangentially at the rim of the pulley. The torque from the downward tension is $T \times r$ (clockwise), and the torque from the horizontal tension is $T \times r$ (counter-clockwise). They perfectly cancel each other out, ensuring the pulley remains in rotational equilibrium.

Q: Can't we just add the masses $M$ and $m$ together and multiply by $g$?

No, because the forces are not all pointing in the same direction! The tension from the wall acts horizontally, while the weights act vertically. Force is a vector, so you must use vector addition (the Pythagorean theorem) to find the true total force.

Q: Why is the clamp force exactly equal to the resultant force $R$?

According to Newton's First Law, for an object to be in static equilibrium (not accelerating), the net force on it must be exactly zero. The clamp is the only thing holding the pulley up, so it must provide a force that perfectly cancels out the combined effect (resultant) of gravity and the string's tension.

Q: What if the string was pulled at an angle instead of perfectly horizontally?

If the string were pulled at an angle $\theta$, you would need to resolve that angled tension into horizontal ($T \cos\theta$) and vertical ($T \sin\theta$) components before summing the forces. The overall method of finding $F_x$, $F_y$, and then the resultant remains exactly the same!

Question

A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by (2001)

Options:

(A) $\sqrt{2} Mg$

(B) $\sqrt{2} mg$

(C) $g \sqrt{(M + m)^2 + m^2}$

(D) $g \sqrt{(M + m)^2 + M^2}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are tasked with holding a heavy bucket using a rope that passes over a pulley. The pulley makes it easier to lift, but have you ever wondered about the immense stress placed on the mount holding the pulley itself?

In this classic physics problem, we are exploring exactly that. We have a block of mass $M$ hanging from a string. This string passes over a pulley of mass $m$, which is firmly attached to a wall by a clamp.

Our mission is to determine the exact force that the clamp must exert on the pulley to prevent the entire system from collapsing. To do this, we must systematically break down the forces acting on each component.

### The Master Equation: Equilibrium of the Block

Let's begin with the easiest part of the system: the hanging block.

Since the block is perfectly at rest, it is in a state of static equilibrium. According to Newton's First Law, the net force acting on it must be zero.

There are only two forces acting on this block. Gravity pulls it downwards with a force equal to its weight, $Mg$. Simultaneously, the string pulls it upwards with a tension force, $T$.

Because these forces must perfectly balance each other, we can write our first crucial equation:
$$T = Mg$$

### The Pulley's Free Body Diagram

Now, we shift our focus to the star of the show: the pulley. To find the force exerted by the clamp, we must isolate the pulley and identify every single force acting upon it.

First, we have the forces from the string. The string wraps around the pulley, pulling on it in two distinct directions. It pulls downwards with tension $T$ (due to the hanging block), and it pulls horizontally to the left with tension $T$ (where it is anchored to the wall).

Second, we must not forget a common trap! The pulley is not massless; it has a mass $m$. Therefore, gravity pulls the pulley itself downwards with a force $mg$.

### Vector Addition: Finding the Resultant

Since force is a vector quantity, we cannot simply add these values together algebraically. We must sum them along their respective axes.

Let's calculate the total vertical force, $F_y$, acting downwards on the pulley. We have the tension $T$ and the pulley's weight $mg$. 
$$F_y = T + mg$$

Substituting our earlier finding that $T = Mg$, we get:
$$F_y = Mg + mg = (M+m)g$$

Next, let's look at the horizontal direction. The only force acting horizontally is the tension $T$ pulling to the left. 
$$F_x = T = Mg$$

Now, we find the resultant active force $R$ acting on the pulley by combining these perpendicular components using the Pythagorean theorem:
$$R = \sqrt{F_x^2 + F_y^2}$$

Substituting our expressions for $F_x$ and $F_y$:
$$R = \sqrt{(Mg)^2 + ((M+m)g)^2}$$

Factoring out the common $g^2$ from under the square root, we arrive at the magnitude of the resultant force:
$$R = g \sqrt{M^2 + (M+m)^2}$$

### Newton's Third Law and the Clamp's Response

We have found the total force trying to rip the pulley off the wall. But the pulley isn't moving!

For the pulley to remain in static equilibrium, the clamp must fight back. It must exert a force $F_c$ that perfectly cancels out the resultant force $R$.

Therefore, the clamp force must be equal in magnitude and exactly opposite in direction to $R$.

**$$F_c = g \sqrt{(M+m)^2 + M^2}$$**

This elegant expression reveals exactly how much strength the clamp needs to hold the entire system together, accounting for both the payload and the hardware itself.

Animated Solution for Physics - Laws of Motion: A string of negligible mass going over a clamped pulley of mass $m$ supports a block of mass $M$ as shown in the figure. The force on the pulley by the clamp is given by (2001)

Select Answer:

Visualized Solution

The Physical Setup

Equilibrium of Block $M$

Isolating the Pulley

Forces from the String

Weight of the Pulley

Total Vertical Force $F_{y}$

Total Horizontal Force $F_{x}$

The Resultant Active Force $R$

Calculating the Resultant

The Clamp Force $F_{c}$

The Sigma Insight: Equilibrium of Concurrent Forces

Analyzing the Setup

The Master Equation

The Pulley's Free Body Diagram

Vector Addition

Newton's Third Law and the Clamp's Response

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