Mechanical Engineering

[nycmathguy]

Section 6.2

Can you do parts (a) & (b)?

 

[MathLover1]

(a) Use the Law of Cosines to write an equation giving the relationship between x and θ

b=7in
a=1.5in
c=x

b^2=a^2+c^2-2ac*cos(θ)
7^2=1.5^2+x^2-2*1.5*x*cos(θ)
49=2.25+x^2-3x*cos(θ)

(b) Write x as a function of θ (Select the sign that yields positive values of x. )

49=2.25+x^2-3x*cos(θ)
0=2.25+x^2-3x*cos(θ)-49
x^2-3cos(θ)*x-46.75=0

x=(-(-3cos(θ)+-sqrt((-3cos(θ))^2-4*1*(-46.75))/(2*1)

select positive solution:
x=(3cos(θ)+sqrt(9cos^2(θ)+187)/2

 

[nycmathguy]

(a) Use the Law of Cosines to write an equation giving the relationship between x and θ

b=7in
a=1.5in
c=x

b^2=a^2+c^2-2ac*cos(θ)
7^2=1.5^2+x^2-2*1.5*x*cos(θ)
49=2.25+x^2-3x*cos(θ)

(b) Write x as a function of θ (Select the sign that yields positive values of x. )

49=2.25+x^2-3x*cos(θ)
0=2.25+x^2-3x*cos(θ)-49
x^2-3cos(θ)*x-46.75=0

x=(-(-3cos(θ)+-sqrt((-3cos(θ))^2-4*1*(-46.75))/(2*1)

select positive solution:
x=(3cos(θ)+sqrt(9cos^2(θ)+187)/2

Thank you. I will endeavor to do the remaining parts.

 

[nycmathguy]

(a) Use the Law of Cosines to write an equation giving the relationship between x and θ

b=7in
a=1.5in
c=x

b^2=a^2+c^2-2ac*cos(θ)
7^2=1.5^2+x^2-2*1.5*x*cos(θ)
49=2.25+x^2-3x*cos(θ)

(b) Write x as a function of θ (Select the sign that yields positive values of x. )

49=2.25+x^2-3x*cos(θ)
0=2.25+x^2-3x*cos(θ)-49
x^2-3cos(θ)*x-46.75=0

x=(-(-3cos(θ)+-sqrt((-3cos(θ))^2-4*1*(-46.75))/(2*1)

select positive solution:
x=(3cos(θ)+sqrt(9cos^2(θ)+187)/2

To graph part (b), I expresses the function as y in terms of x where x = theta. I don't have a theta symbol on my keyboard setting.

For part (d), the piston moves 5 inches in one cycle.

 

[nycmathguy]

(a) Use the Law of Cosines to write an equation giving the relationship between x and θ

b=7in
a=1.5in
c=x

b^2=a^2+c^2-2ac*cos(θ)
7^2=1.5^2+x^2-2*1.5*x*cos(θ)
49=2.25+x^2-3x*cos(θ)

(b) Write x as a function of θ (Select the sign that yields positive values of x. )

49=2.25+x^2-3x*cos(θ)
0=2.25+x^2-3x*cos(θ)-49
x^2-3cos(θ)*x-46.75=0

x=(-(-3cos(θ)+-sqrt((-3cos(θ))^2-4*1*(-46.75))/(2*1)

select positive solution:
x=(3cos(θ)+sqrt(9cos^2(θ)+187)/2

Is my answer for parts (c) & (d) correct?

 

[MathLover1]

yes

 

[nycmathguy]

yes

Perfect.