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2 years ago

Determine the cutting force f exerted on the rod s in terms of the forces p applied to the handles of the heavy-duty cutter.

2 answers:

6 0

CASE 1: Direction vertically downward on the blade

Assume pin force at E acts downward on the blade and CW is a positive moment 
Ey[3.4] - F[1.7] = 0 
Ey = F/2 

Case 2: Direction vertically upward on the blade . Assume CW is positive moment 

Ey[1.5sin19] – P[21 – 1.5sin19] = 0 
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

Alexus [3.1K]2 years ago

5 0

If you use the next formula with the data given in the exercise you are asking:
Ey[3.4] - F[1.7] = 0
Ey = F/2
and after that what you need to do is sum the moments of the handle about D to zero asumming it is a positive moment and we proceed like this
Ey[1.5sin19] – P[21 – 1.5sin19] = 0
(F/2)[1.5sin19] = P[21 – 1.5sin19] 
F = 2P[21 – 1.5sin19] / [1.5sin19] 
F = 84P

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An engineer wants to design a circular racetrack of radius R such that cars of mass m can go around the track at speed without t

gtnhenbr [62]

1. $tan \theta = \frac{v^2}{Rg}$

For the first part, we just need to write the equation of the forces along two perpendicular directions.

We have actually only two forces acting on the car, if we want it to go around the track without friction:

- The weight of the car, mg, downward

- The normal reaction of the track on the car, N, which is perpendicular to the track itself (see free-body diagram attached)

By resolving the normal reaction along the horizontal and vertical direction, we find the following equations:

$N cos \theta = mg$ (1)

$N sin \theta = m \frac{v^2}{R}$ (2)

where in the second equation, the term $m\frac{v^2}{R}$ represents the centripetal force, with v being the speed of the car and R the radius of the track.

Dividing eq.(2) by eq.(1), we get the following expression:

$tan \theta = \frac{v^2}{Rg}$

2. $F=\frac{m}{R}(w^2-v^2)$

In this second situation, the cars moves around the track at a speed

$w>v$

This means that the centripetal force term

$m\frac{v^2}{R}$

is now larger than before, and therefore, the horizontal component of the normal reaction, $N sin \theta$ , is no longer enough to keep the car in circular motion.

This means, therefore, that an additional radial force F is required to keep the car round the track in circular motion, and therefore the equation becomes

$N sin \theta + F = m\frac{w^2}{R}$

And re-arranging for F,

$F=m\frac{w^2}{R}-N sin \theta$ (3)

But from eq.(2) in the previous part we know that

$N sin \theta = m \frac{v^2}{R}$

So, susbtituting into eq.(3),

$F=m\frac{w^2}{R}-m\frac{v^2}{R}=\frac{m}{R}(w^2-v^2)$

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1 year ago

Which of the following sketches represents a possible configuration for this problem?

garri49 [273]

Where are the following sketches?

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2 years ago

Suggest reasons why poaching for subsistence is likely to be less damaging to the biodiversity of an area than poaching for prof

dlinn [17]

An example for ruining a biodiversity is fishing. The two factors that have contributed to increased fishing in deep ocean waters in recent years are the human population growth and decreased fishing opportunities inshore. Increase population growth increases the demand for food which also leads to increase in fish demand. Because the fish demand is high, inshore fishing opportunities decrease that is why deep ocean waters is the new venue for fishing. This may sound absurd but poaching for subsistence is likely to be less damaging to he biodiversity of an area than poaching for profit. Because the people do not care anymore to the biodiversity that they interrupted just to get back more profit. They do not care what must be taken from it like getting bigger fishes and leaving the smaller ones behind to maintain productivity.

5 0

2 years ago

Paul and Ivan are riding a tandem bike together. They’re moving at a speed of 5 meters/second. Paul and Ivan each have a mass of

Sonbull [250]

The formula is Ke = 1/2 m v^2
The two of them together have a Ke of mv^2. So you either increase m or v. That's what makes the problem difficult. He can do D or B. We have to choose.

A is no solution. The Ke goes down because Paul loses Ivan's mass.
C is out of the question 3 meters/sec is a big reduction from 5 m/s. So now what do we do about B and D?

The question is what does the third person add. The tandoms I've peddled only allow for 1 or 2 people to add to the motion. So the third person only adds mass. He does not have a v that he is contributing to. To say that he is going 5m/s is true, but he's not contributing anything to that motion.

I pick B, but it is one of those questions that the correctness of it is in the head of the proposer. Be prepared to get it wrong. Argue the point politely if you agree with me, but back off as soon as you have presented your case.

B <<<<====== answer.

5 0

2 years ago

Read 2 more answers

Two cars start 200 m apart and drive toward each other at a steady 10 m/s. On the front of one of them, an energetic grasshopper

vladimir1956 [14]

Answer:

Total distance does the grasshopper travel before the cars hit is 150 m

Explanation:

Each car moves x=100 m before they collide. Both the cars moving in constant velocity. time taken t by each car is

$t=\frac{x}{v}$