A solid uniform ball with mass m and diameter d is supported against a vertical frictionless wall by a thin massless wire of length L. a) Find the tension in the wire.Let us first derive the expression for the tension.By Newton's First LawΣF = the components of force that makes this so isΣF = T - = 0Simplifying this results toT = g/(ϕ)Analyzing the figure above, we arrive at(ϕ) = sqrt( 2 + ) / ( / + )By substitution, we arrive at the following:T = ( + 2 ) g / ( sqrt ( 2 + ) )If the ball has a mass of 45 kg and diameter 32 cm, while the wire has a length of 30 cm. The tension is equal toT = 0.370 N Solutiona)Let us first derive the expression for the tension.By Newton's First LawEF = 0the components of force that makes this so isIF = T costetha= 0mgSimplifying this results toT= mg/ cosAnalyzing the figure above, we arrive at(4) = sqrt(+cosBy substitution, we arrive at the following:T= m+ 2sqrt (2+))If the ball has a mass of 45 kg and diameter 32 cm, while the wire has a length of 30 cm. The tension is equal toT=0.370 N2.+ A solid uniform ball with mass m and diameter d is supported against a vertical frictionless wall by a thin massless wire of length L. a) Find thetension in the wire.Ld

Using Newton's law, ∑F=0 From the above expressions, Tcosϕ-mg=0 Simplifying this results to T=mgcosϕ…

a) Let us first derive the expression for the tension. By Newton's First Law EF = 0 the components of force that makes this so is EF = T costetha mg = 0 Simplifying this results to T= m g/ cos Analyzing the figure above, we arrive at 2+ cos (0) = sqrt( By substitution, we arrive at the following: T= m + 2 sqrt ( )) If the ball has a mass of 45 kg and diameter 32 cm, while the wire has a length of 30 cm. The tension is equal to T = 0.370 N 2.

a) Let us first derive the expression for the tension. By Newton's First Law EF = 0 the components of force that makes this so is EF = T costetha mg = 0 Simplifying this results to T= m g/ cos Analyzing the figure above, we arrive at 2+ cos (0) = sqrt( By substitution, we arrive at the following: T= m + 2 sqrt ( )) If the ball has a mass of 45 kg and diameter 32 cm, while the wire has a length of 30 cm. The tension is equal to T = 0.370 N 2.

University Physics Volume 1

18th Edition

ISBN:9781938168277

Author:William Moebs, Samuel J. Ling, Jeff Sanny

Publisher:William Moebs, Samuel J. Ling, Jeff Sanny

Chapter12: Static Equilibrium And Elasticity

Section: Chapter Questions

Problem 59P: A copper wire is 1.0 m long and it diameter is 1.0 mm. if the wire hangs vertically how much weight...

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Concept explainers

Mechanical Properties of Solids

There are three types of matter- solid, liquid and gas. Out of them, solids have a rigid shaped and size. The intermolecular forces of attractions in solids are greatest. Mechanical properties are the physical properties of a material which it shows, when it is provided with certain amount of force or load. These properties are to be kept in mind for various purposes. When a force is applied to an object, the object may or may not get deformed depending upon its elastic property. But every time the object develops a restoring force against the force applied. This is commonly known as stress. If the object changes its dimensions due to the force applied it develops strain ( which is the ratio of change in dimension to the original dimension).

Question

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A solid uniform ball with mass m and diameter d is supported against a vertical frictionless wall by a thin massless wire of length L. a) Find the tension in the wire.

Let us first derive the expression for the tension.

By Newton's First Law

ΣF =

the components of force that makes this so is

ΣF = T - = 0

Simplifying this results to

T = g/(ϕ)

Analyzing the figure above, we arrive at

(ϕ) = sqrt( ² + ) / ( / + )

By substitution, we arrive at the following:

T = ( + 2 ) g / ( sqrt ( ² + ) )

If the ball has a mass of 45 kg and diameter 32 cm, while the wire has a length of 30 cm. The tension is equal to

T = 0.370 N

Solution
a)
Let us first derive the expression for the tension.
By Newton's First Law
EF = 0
the components of force that makes this so is
IF = T costetha
= 0
mg
Simplifying this results to
T= m
g/ cos
Analyzing the figure above, we arrive at
(4) = sqrt(
+
cos
By substitution, we arrive at the following:
T= m
+ 2
sqrt (
2+
))
If the ball has a mass of 45 kg and diameter 32 cm, while the wire has a length of 30 cm. The tension is equal to
T=
0.370 N
2.
+

A solid uniform ball with mass m and diameter d is supported against a vertical frictionless wall by a thin massless wire of length L. a) Find the
tension in the wire.
L
d

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