In Fig. 12-60, a 103 kg uniform log hangs by two steel...

In Fig. $12-60,$ a 103 kg uniform log hangs by two steel wires, $A$ and $B,$ both of radius 1.20 $\mathrm{mm}$ . Initially, wire $A$ was 2.50 $\mathrm{m}$ long and 2.00 $\mathrm{mm}$ shorter than wire $B$ . The log is now horizontal. What are the magnitudes of the forces on it from (a) wire $A$ and (b) wire $B ?(\mathrm{c})$ What is the ratio $d_{A} / d_{B} ?$

In Fig. $12-60,$ a 103 kg uniform log hangs by two steel wires,
$A$ and $B,$ both of radius 1.20 $\mathrm{mm}$ .
Initially, wire $A$ was 2.50 $\mathrm{m}$ long
and 2.00 $\mathrm{mm}$ shorter than wire $B$ .
The log is now horizontal. What
are the magnitudes of the forces
on it from (a) wire $A$ and (b) wire
$B ?(\mathrm{c})$ What is the ratio $d_{A} / d_{B} ?$

Key Concepts

Static Equilibrium

Static equilibrium is the state in which all forces and torques acting on a system sum to zero. This principle ensures that objects remain at rest or in constant motion, and it is fundamental when analyzing support systems where multiple forces, such as gravitational forces and tension forces, must balance each other.

Force Distribution in Parallel Supports

When a load is supported by more than one element, each support carries a portion of the load depending on its stiffness. In systems with parallel supports, the forces are distributed according to each element’s resistance to deformation, meaning that supports that stretch less under load typically bear a larger share of the weight.

Elastic Deformation and Hooke's Law

Elastic deformation is the behavior of materials under a load when they deform reversibly. Hooke’s law states that, within the elastic limit, the extension of an object is directly proportional to the force applied. This concept is essential for analyzing how supports that stretch under tension will respond to varying forces.

Young's Modulus

Young's modulus is a material property that defines the relationship between stress (force per unit area) and strain (relative deformation). It is a measure of the stiffness of a material; higher values indicate a stiffer material that deforms less under an applied load. This concept is key when calculating elongation in elastic members such as wires or rods.

Transcript

00:01 For this problem on the topic of equilibrium and elasticity, we are given a 103 kilogram uniform log that is hanging by two steel wires a and b.

00:10 Both wires have a radius of 1 .2 millimeters and initially wire a is 2 .5 meters long and 2mm shorter than wire b.

00:20 The log is now in a horizontal position and we want to know the magnitudes of the forces that wire a exert on it and wire b exert on it.

00:29 We will lastly want to calculate the ratio of distances da to db.

00:35 So firstly we let fa and fb be the forces exerted by the wires in the log and let m be the mass of the log.

00:42 Since the log is in equilibrium, the force that a exerts in the log, that's the force that b exerts in the log minus the weight of the log mg must equal to zero.

00:54 Now information given about the stretching of the wires will allow us to find the relationship between fa and fb.

01:00 So if wire a originally had a length laa and stretches by a distance delta laa which is its extension then we have that this extension for a delta la is equal to be four supplied on the log by wire a times its original length la over its cross -sectional area a times the young's modulus for steel which is 200 times 10 to the power 9 newtons per square meter.

01:35 Now we can do the same for the extension of wire b.

01:38 Delta lb is fb times lb over the cross -sectional area which is the same as wire a since it has the same radius multiplied again by the same young's modulus which is the young's modulus for steel.

01:54 Now if we let the value little l be the amount by which b was originally longer than a and since they have the same length after the log is attached, we have that delta la is equal to delta lb plus little l.

02:15 And so this means from the equations above that f -a -l -a over a times e is equal to f -b -l -b over a times e plus little l.

02:33 So we can rearrange this equation and we can solve for fb.

02:38 And we get fb is equal to f -a -l -a over l -b minus a -e -l -l over l -b...

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