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The following article will add knowledge about “** One end of a uniform meter stick is placed against a vertical wall as shown in**“. Let’s not forget to look at the content!

## Question

One end of a uniform meter stick is placed against a vertical wall as shown in

One end of a uniform meter stick is placed against a vertical wall as shown in (Figure 1) . The other end is held by a lightweight cord that makes an angle θ with the stick. The coefficient of static friction between the end of the meter stick and the wall is 0,40.

What is the maximum value the angle θ can have if the stick is to remain in equilibrium?

Let the angle between the cord and the stick is θ = 14∘. A block of the same weight as the meter stick is suspended from the stick, as shown at (Figure 2), at a distance x from the wall. What is the minimum value of x for which the stick will remain in equilibrium?

When θ = 14∘, how large must the coefficient of static friction be so that the block can be attached 14 cm from the left end of the stick without causing it to slip?

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## Answer “One end of a uniform meter stick is placed against a vertical wall as shown in”.

### The Principle of Moments:

If we select a point in equilibrium for an object, the algebraic sum and torque of that object around the point is zero. This principle can be used to balance the forces horizontally and vertically.

### Torque:

A force that causes an object to turn upon a fixed axis point, also known as a pivot, is called torque. The torque that an object experiences will equal the force applied multiplied by the length of the arm causing it to spin will be the torque. An object will experience angular acceleration when it experiences torque.

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**Answer and Explanation:**

**Part 1: **

Before we can determine the angle that the cord makes with meterstick, it is necessary to determine the value of tension that the cord experiments. The system is now in equilibrium. We can see that all forces parallel to the wall should be equal.

ƩF_{y }= 0

fr − m × g + Tsinθ = 0

μ_{s}Fn – m × g + Tsinθ = 0

We don’t know how much the normal force is worth so we need to analyze forces perpendicular the wall:

_{ }ƩF_{x} = 0

F_{n}−Tcosθ = 0

F_{n} = Tcosθ

Substituting in the previous equation:

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There are 2 more unknowns. We are analyzing the torques and working on the system that defines the pivot at pin.

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Substituting in our current expression:

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**Part 2:**

We have a new condition, and we have a different angle for the cord. We also determine the relationship between tension and weight. We first analyze the equilibrium perpendicularly to the wall.

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Analyzing the forces parallel to the wall:

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We then calculate the distance x by analysing the torques used in the system. We again define the pivot at pin.

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**Part 3:**

The friction coefficient is the final part of the problem. We use torque this time to determine the relationship between the system’s weight and the tension in our cord.

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Analyzing the forces perpendicular to the wall:

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Lastly, let’s examine the forces that are parallel to the wall.

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## Last words

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