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skad [1K]
3 years ago
5

A mechanism is a device that transmits movement so the output movement is different from the input movement. identify three ways

that the output movement could be different from the input
Help Me out !!
Engineering
1 answer:
igomit [66]3 years ago
7 0

Answer:

• Gear ratios compare the output (or driven gear)

to the input (or drive gear)

• Gear Ratios can be determined using number

(n) of teeth on the gear or diameter (d) of the

gear

• If the output gear is larger than the

input gear the speed will decrease

• If the output gear is smaller than the

input gear the speed will increase

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Determine the magnitude of the resultant force and the moment about the origin. Note: the symbol near the 140 N-m moment are not
arlik [135]

Answer:

R = 148.346 N

M₀ = - 237.2792 N-m

Explanation:

Point O is selected as a convenient reference point for the force-couple system which is to represent the given system

We can apply

∑Fx = Rx = - 60N*Cos 45° + 40N + 80*Cos 30° = 66.8556 N

∑Fy = Ry = 60N*Sin 45° + 50N + 80*Sin 30° = 132.4264 N

Then

R = √(Rx²+Ry²)    ⇒  R = √((66.8556 N)²+(132.4264 N)²)

⇒  R = 148.346 N

Now, we obtain the moment about the origin as follows

M₀ = (0 m*40 N)-(7 m*60 N*Sin 45°)+(4 m*60 N*Cos 45°)-(5 m*50 N)+ 140 N-m + (0 m*80 N*Cos 30°) + (0 m*80 N*Sin 30°) = - 237.2792 N-m (clockwise)

We can see the pic shown in order to understand the question.

7 0
3 years ago
Tech A says that in some cases, the electronic brake control module can be programmed with a new tire size to restore proper ele
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Both Techs A and B

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8 0
2 years ago
The state of plane strain on an element is:
balu736 [363]

Answer:

a. ε₁=-0.000317

   ε₂=0.000017

θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain =3.335 *10^-4

Associated average normal strain ε(avg) =150 *10^-6

θ = 31.71 or -58.29

Explanation:

\epsilon _{1,2} =\frac{\epsilon_x + \epsilon_y}{2}  \pm \sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\epsilon _{1,2} =\frac{-300 \times 10^{-6} + 0}{2}  \pm \sqrt{(\frac{-300 \times 10^{-6}+ 0}{2}) ^2 + (\frac{150 \times 10^-6}{2})^2}\\\\\epsilon _{1,2} = -150 \times 10^{-6}  \pm 1.67 \times 10^{-4}

ε₁=-0.000317

ε₂=0.000017

To determine the orientation of ε₁ and ε₂

tan 2 \theta_p = \frac{\gamma_xy}{\epsilon_x - \epsilon_y} \\\\tan 2 \theta_p = \frac{150 \times 10^{-6}}{-300 \times 10^{-6}-\ 0}\\\\tan 2 \theta_p = -0.5

θ= -13.28° and  76.72°

To determine the direction of ε₁ and ε₂

\epsilon _{x' }=\frac{\epsilon_x + \epsilon_y}{2}  + \frac{\epsilon_x -\epsilon_y}{2} cos2\theta  + \frac{\gamma_xy}{2}sin2\theta \\\\\epsilon _{x'} =\frac{-300 \times 10^{-6}+ \ 0}{2}  + \frac{-300 \times 10^{-6} -\ 0}{2} cos(-26.56)  + \frac{150 \times 10^{-6}}{2}sin(-26.56)\\\\

=-0.000284 -0.0000335 = -0.000317 =ε₁

Therefore θ₁= -13.28° and  θ₂=76.72°  

b. maximum in-plane shear strain

\gamma_{max \ in \ plane} =2\sqrt{(\frac{\epsilon_x + \epsilon_y}{2} )^2 + (\frac{\gamma_xy}{2})^2} \\\\\gamma_{max \ in \ plane} = 2\sqrt{(\frac{-300 *10^{-6} + 0}{2} )^2 + (\frac{150 *10^{-6}}{2})^2}

=3.335 *10^-4

\epsilon_{avg} =(\frac{\epsilon_x + \epsilon_y}{2} )

ε(avg) =150 *10^-6

orientation of γmax

tan 2 \theta_s = \frac{-(\epsilon_x - \epsilon_y)}{\gamma_xy} \\\\tan 2 \theta_s = \frac{-(-300*10^{-6} - 0)}{150*10^{-6}}

θ = 31.71 or -58.29

To determine the direction of γmax

\gamma _{x'y' }=  - \frac{\epsilon_x -\epsilon_y}{2} sin2\theta  + \frac{\gamma_xy}{2}cos2\theta \\\\\gamma _{x'y' }=  - \frac{-300*10^{-6} - \ 0}{2} sin(63.42)  + \frac{150*10^{-6}}{2}cos(63.42)

= 1.67 *10^-4

4 0
3 years ago
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