A mobile is formed by supporting four metal butterflies of...

A mobile is formed by supporting four metal butterflies of equal mass $m$ from a string of length $L$ . The points of support are evenly spaced a distance $\ell$ apart as shown in Figure $\mathrm{P} 5.73$ . The string forms an angle $\theta_{1}$ with the ceiling at each end point. The center section of string is horizontal. (a) Find the tension in each section of string in terms of $\theta_{1}, m,$ and $g .$ (b) Find the angle $\theta_{2},$ in terms of $\theta_{1},$ that the sections of string between the outside butterflies and the inside butterfles form with the horizontal. (c) Show that the distance $D$ between the end points of the string is $$ D=\frac{L}{5}\left(2 \cos \theta_{1}+2 \cos \left[\tan ^{-1}\left(\frac{1}{2} \tan \theta_{1}\right)\right]+1\right) $$

A mobile is formed by supporting four metal butterflies of equal mass $m$ from a string of length $L$ . The points of support are evenly spaced a distance $\ell$ apart as shown in Figure $\mathrm{P} 5.73$ . The string forms an angle $\theta_{1}$ with the ceiling at each end point. The center section of string is horizontal. (a) Find the tension in each section of string in terms of $\theta_{1}, m,$ and $g .$ (b) Find the angle $\theta_{2},$ in terms of $\theta_{1},$ that the sections of string between the outside butterflies and the inside butterfles form with the horizontal. (c) Show that the distance $D$ between the end points of the string is
$$
D=\frac{L}{5}\left(2 \cos \theta_{1}+2 \cos \left[\tan ^{-1}\left(\frac{1}{2} \tan \theta_{1}\right)\right]+1\right)
$$

Key Concepts

Geometric Analysis of Structures

Geometric analysis involves understanding the spatial arrangement of components in a physical system. In the context of mobiles or similar structures, analyzing the relative lengths and angles allows one to relate the spatial configuration of the system to the forces acting on it. This approach is key for determining distances, angles between structural elements, and how the distribution of mass influences the overall balance and stability of the system.

Trigonometric Relationships in Mechanics

Trigonometry is used to relate angles to the ratios of sides in right triangles, which is especially relevant when analyzing systems with inclined members such as strings or beams. Understanding how to use functions like tangent, sine, and cosine is critical for expressing the geometric relationships between various parts of a structure and for converting between forces acting along different directions.

Force Resolution

Force resolution refers to breaking a force into its components along chosen coordinate axes, typically horizontal and vertical. In scenarios involving inclined strings or cables, this process uses trigonometric functions such as sine and cosine to relate the magnitude of the force to its horizontal and vertical components, enabling the application of equilibrium equations in each direction.

Static Equilibrium

This concept involves ensuring that all forces and moments (torques) acting on a system are balanced so that the system remains at rest. In problems dealing with structures or hanging objects, applying the conditions of static equilibrium (i.e., the net force in every direction is zero, and the net moment about any point is zero) is fundamental for setting up and solving equations to find unknown forces or tensions.

Transcript

00:01 Okay, so string forms a mobile as shown in the diagram with butterfly with equal mass attached in a string of total length of l, okay, and it is attached at equal distance of length l, and it forms an angle, theta 1 from the ceiling at both the ends and theta 2 here as shown and therefore it is symmetric.

00:20 Okay, so diagram is symmetric or mobile is symmetric.

00:23 So now we can apply newton's second law of motion at every node, okay, to solve.

00:29 So at first node, we'll have mass acting downwards then we have this let's take as t2 and this is this t1 okay now this is making an angle of theta 1 here and this is making an angle of theta 2 so on balancing the force that is we get vertical force will be so let me draw it bigger so that we can see what is happening at that point so the point let's say this is a b right c and d will be same as a and b so we'll all for a and b.

01:00 So point a we have mass mg acting downwards.

01:04 Then we have a cable or string tension acting this way with angle theta 2 and then we have another string tension.

01:14 Let's take a st2, this one is t1 and it is angle of theta 1.

01:18 Okay so summation of fx must be equals to 0.

01:22 Okay so here we can divide this t1 into or into components horizontal and vertical.

01:28 So our horizontal component is t1 cost theta 1 and which is in negative so minus plus this t2 can be here in this direction and downward direction so we have t2 cost theta 2 in right direction and this must be equal to 0 so let this equation 1.

01:49 Similarly summation of f y must be equal to 0 so here t1 is t1 sine theta is active upwards so we'll take a positive t1 sine theta 1 and t1 sine theta 2 acting downwards so minus t 2 sine theta 2 and this must be equals to the applied low so that is equal to m g or minus m g must be equal to 0 so this is a second equation on similar basis we can solve for 3 and 4 that is at this point sorry at this point b okay so at solving for point b we get since t3 is horizontal right so let me solve it so here point b we have here t3 mass mg and we have no t2.

02:33 This t2 is at angle of theta 2 right? so this is theta 2.

02:38 So again summation of fx is equal to 0.

02:41 This t2 can be factor into two components horizontal and vertical.

02:44 So this is t2 cost theta 2 negative plus t3 which is positive right right? right for direction this is equal to 0.

02:52 Equation 3 and summation of f y is 0 so that is here we have t 2 sine theta 2 and it is balanced by m g right in negative direction which is equal to 0 so that is equation 4 great okay so we have four equations and 4 unknown t 1 t2 theta 1 and theta 2 okay so on substituting 4 in 2 1 1 1 2 1 sorry 2 okay so we'll substitute 4 into 2 we get t 1 sine theta 1 minus m g minus m g right this is t 2 sine theta 2 2 yeah t 2 sin theta 2 is minus m g right that is equal to m g so this is equal to 0 and therefore we get t 1 as 2 m g over sine theta 1 great now if you substitute 3 in 1 okay so we get t 3 minus t 1 x -teta 1 equals to 0 so that is t3 is equal to t 1 theta 1 okay so on solving this we get t 3 as 2mg cross theta 1 upon sine theta 1 and therefore t 3 is 2mg upon tan theta 1 great now from equation 4 so directly we have t2 is mg upon sine theta 2.

04:26 Now dividing, okay, so this is done right, we have all the values t1, t2 and t3.

04:33 Okay, so now we'll solve for part b...

Please give Ace some feedback