Please help explain the annotation? FBD for half the pliers (front lever)b = 38 mmPin== 090°Therefore,α-40°15 mmRy50°From geometry, we haveRx50 mmConsider the forces acting on the front lever. R, and R, are the componentsof forces from the pin along the x and y axes, respectively.140°According to the moment equilibrium about the pin, we haveC(b)ΣMpinC(b)- P(a) = 0a125 mmAccording to the force equilibrium along the x axis, we haveΣΕ = 0>Ccos(a) + Rx = 0Rx = -Ccos(a)According to the force equilibrium along the y axis, we haveΣF₁ = 0R₁ - P-Csin(a) = 0y90⁰P =R₁ = P +Csin(a)=C(=+sin(a))) the byaThe magnitude of the resultant force acting on the pin isabR = √ R² + R² Cx)√ (cos(a))² + (−+sin(a))²{cxaR = Cx₁ (cos(40°))² + (;= 1.163× C38mm163.3mmor C=a = 125mm +50mm× cos(a) = 125mm +50mm× cos(40°)=163.3mm-How? specifically+ sin(40°))P(a)bwhy?How is this 40°

Answered: Image /qna-images/answer/b761fde8-90b4-408b-b666-b35f1e513745.jpg

FBD for half the pliers (front lever) b = 38 mm Pin 90° α-40° 15 mm Ry 50° Rx From geometry, we have 50 mm 140° 125 mm P Consider the forces acting on the front lever. R. and R, are the components of forces from the pin along the x and y axes, respectively. According to the moment equilibrium about the pin, we have ΣMpin = 0 C(b)- P(a) = 0 C(b) a 10° 90° P= = According to the force equilibrium along the x axis, we have ΣΕ = 0 >Ccos(a)+Rx = 0 Rx = -Ccos(a) According to the force equilibrium along the y axis, we have ΣF = 0 Ry -P-Csin(a)= 0 b R₁ = P + Csin(a)=C(−+sin(a)) a The magnitude of the resultant force acting on the pin is Therefore, R=Cx₁ (cos(40°))² + (; 38mm 163.3mm b R = √R² + R²} {Cx) {(cos(a))² + ( + sin(a))² a or C= How? specifically the b? a a = 125mm +50mm× cos(a) = 125mm +50mmx cos(40°) =163.3mm + sin(40°)) P(a) b why? How is this 40°

FBD for half the pliers (front lever) b = 38 mm Pin 90° α-40° 15 mm Ry 50° Rx From geometry, we have 50 mm 140° 125 mm P Consider the forces acting on the front lever. R. and R, are the components of forces from the pin along the x and y axes, respectively. According to the moment equilibrium about the pin, we have ΣMpin = 0 C(b)- P(a) = 0 C(b) a 10° 90° P= = According to the force equilibrium along the x axis, we have ΣΕ = 0 >Ccos(a)+Rx = 0 Rx = -Ccos(a) According to the force equilibrium along the y axis, we have ΣF = 0 Ry -P-Csin(a)= 0 b R₁ = P + Csin(a)=C(−+sin(a)) a The magnitude of the resultant force acting on the pin is Therefore, R=Cx₁ (cos(40°))² + (; 38mm 163.3mm b R = √R² + R²} {Cx) {(cos(a))² + ( + sin(a))² a or C= How? specifically the b? a a = 125mm +50mm× cos(a) = 125mm +50mmx cos(40°) =163.3mm + sin(40°)) P(a) b why? How is this 40°

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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Concept explainers

Chemical and Phase Equilibrium

Phase diagrams or equilibrium phase diagrams are the graphical representations that show the phase transformations of the constituting elements with respect to the pressure, temperature, and compositions. The diagrams show the conditions that limit the different phases, namely solid, liquid, and gas. The diagram can also be utilized to verify the relations between the state variables. Most of the phase diagrams are temperature-pressure phase diagrams, with the variable parameter as the composition. Usually other variables such as solubility and pressure parameters are not indicated in the phase diagram. For a one-component system (unary phase diagram), the phase diagram is generally a function of temperature and pressure as the state variables, or temperature and volume. However, for a binary system (two-phase) having two components, the phase diagram is a function of temperature and composition as the state variables only. Coexisting of the different phases of the constituting elements is indicated in the phase diagram. A ternary phase diagram can be characterized by consisting of three phases in equilibrium.

Question

Please help explain the annotation?

FBD for half the pliers (front lever)
b = 38 mm
Pin
=
= 0
90°
Therefore,
α-40°
15 mm
Ry
50°
From geometry, we have
Rx
50 mm
Consider the forces acting on the front lever. R, and R, are the components
of forces from the pin along the x and y axes, respectively.
140°
According to the moment equilibrium about the pin, we have
C(b)
ΣMpin
C(b)- P(a) = 0
a
125 mm
According to the force equilibrium along the x axis, we have
ΣΕ = 0
>Ccos(a) + Rx = 0
Rx = -Ccos(a)
According to the force equilibrium along the y axis, we have
ΣF₁ = 0
R₁ - P-Csin(a) = 0
y
90⁰
P =
R₁ = P +Csin(a)=C(=+sin(a))) the b
y
a
The magnitude of the resultant force acting on the pin is
a
b
R = √ R² + R² Cx)√ (cos(a))² + (−+sin(a))²
{cx
a
R = Cx₁ (cos(40°))² + (;
= 1.163× C
38mm
163.3mm
or C=
a = 125mm +50mm× cos(a) = 125mm +50mm× cos(40°)
=163.3mm
-How? specifically
+ sin(40°))
P(a)
b
why?
How is this 40°

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