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Scorpion4ik [409]

3 years ago

13

How do i calculate the force acting on the actuator in the suspension system of a fsae car in order to calculate the pressure go

ing in to the accumulator?

1 answer:

Vlad [161]3 years ago

5 0

Answer:

Force acting on the actuator in the suspension system can be measured by the help of a force transducer.

Mostly used sensor are based on piezoelectric quartz crystal or strain gauge sensing elements.We know that sensor is the device which measure the change in the dimension of material by the help of electric circuit.

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Entropy change is evaluated using Eq. 6.2a based on an internally reversible process. Can the entropy change between two states

Vadim26 [7]

Answer:

YES

Explanation:

Entropy is an extensive property of the system entropy change that value of entropy change can be determined for any process between the states whether reversible or not. i have attached the formula to calculate entropy change which is independent of whether the system is reversible or not and can be determined for any process.

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4 years ago

When something is plumb, it is

GrogVix [38]

Answer:

C. exactly vertical

Explanation:

"Plumb" means "exactly vertical."

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4 years ago

Read 2 more answers

A signal expected to be of the form F(t)=10sin15.7t is to be measured with a first-order instrument having a time constant of 50

svet-max [94.6K]

Answer:

The answer is attached below

Explanation:

5 0

4 years ago

Which article of the NEC describes the proper placement of outlets in this type of room? Also, based on the given load what type

zmey [24]

The Article of the NEC that describes the proper placement of outlets in rooms is the Article 210.

<h3>What is NEC?</h3>

NEC is the short or acronym for National Electric Code.

For example, with regard to Receptacles, section (210-21(b)) of the Article 100 indicates that a "Receptacle" is a contact device provided at an outlet for the connecting of an attachment plug.

The terms "receptacle" and "receptacles" occur often in the National Electrical Code (NEC).

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7 0

2 years ago

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these d

Tems11 [23]

Complete Question

Lumber jacks use cranes and giant tongs to hoist their goods into trucks for transport. Fortunately, smaller versions of these devises are available for weekend warriors who want to play with their chain saws. Let us model the illustrated tongs as a planar mechanism that carries a log of weight 210 N. Given the following dimensions: 35 mm 10 mm 40 mm 230 mm 85 mm 45 mm 10 mm 35 mm 345 mm determine the force in N and moment in Nm that our worker exerts on the tongs. Also determine the pinching force magnitude in N that the tongs exert on the log; i.e. determine the horizontal force that the tong's teeth exert on the log. Assume the point E is centered between the tong's teeth.

The diagram for this question is shown on the first uploaded image

Answer:

The force P is $P= 210 N$

and the moment M is $M = -48.3N \cdot m$

The horizontal force that the tong teeth exerts is $F_T =89.67N$

Explanation:

First let denote the dimension to corresponding to the diagram

$a=35mm , b= 10mm, c= 40mm, d= 230mm, e= 85mm,f= 45mm,\\g= 10mm,h=35mm,i=345mm.$

Next looking at the diagram let us consider the vertical direction

At equilibrium

$\sum F_{vertical} =0$

This mean that

$P+ W = 0$

Since they are acting in opposite direction the equation becomes

$P - W = 0$

=> $P= W$

=> $P= 210 N$

At Equilibrium Moment about F gives

$\sum M_f = 0$

=> $F_T * (e +f + g+ h+ i) - F_T * (e+ f+g+ h+i) - W *d -M =0$

=> $M = -W *d$

=> $M = -210 * 0.230$

=> $M = -48.3N \cdot m$

Here $F_T$ is the horizontal force that the tong teeth exerts

Now let consider the part BAF of the system as shown on the second uploaded image

Now the angle $\theta$ is mathematically given as

$tan \theta = \frac{g+h}{a}$

=> $\theta = arctan \frac{g+h}{a}$

$= arctan (\frac{10+35}{35} )$

$=52.125^o$

Now at equilibrium the moment about A is

$\sum M_A = 0$

=> $P * (c+d) +M + F_{BC} cos \theta *f+F_{BC}sin\theta *(a+b) =0$

$210 * (0.040 + 0.230)-48.3+F_{BC} cos (52.125^o)*0.045+ F_{BC}sin(52.125^o)$ $* (0.035 +0.010) =0$

=> $10.29 +F_{BC} (0.02763+0.03552) =0$

=> $F_{BC} =\frac{10.29}{0.06315}$

=> $F_{BC} = - 162.925 N$

Looking at the forces acting on the teeth as shown on the third uploaded image

At Equilibrium the moment about D is

$\sum M_D = 0$

=> $\frac{W}{2} *d - F_T *(i+h) -F_{BC} cos \theta *h -F_{BC} sin \theta * (b+c) =0$

=> $\frac{210}{2} * 0.230 -F_T (0.345 +0.035) - (-162.925)cos(52.125^o) *0.035\\-(-162.925)sin(52.125^o)(0.010 +0.040) =0$

=> $34.081 = F_T(0.345 +0.035)$

=> $F_T =89.67N$

7 0

4 years ago