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ratelena [41]
3 years ago
6

A car is traveling at 50 ft/s when the driver notices a stop sign 100 ft ahead and steps on the brake. Assuming that the deceler

ation due to braking is constant, what is the magnitude of the minimum deceleration for which the car would stop right at the stop sign?
Engineering
1 answer:
Naddika [18.5K]3 years ago
4 0

Answer:

minimum deceleration = 12.5 ft/s²

Explanation:

Data provided in the question:

Initial speed of the car, u  = 50 ft/s

Distance to cover, s = 100 ft

Final speed of the car = 0 ft/s             [ ∵ as the car stops ]

Now,

from the Newton's equation of motion

we have  

v² - u² = 2as

a is the acceleration

on substituting the respective values, we get

0² - 50² = 2 × a × 100

or

- 2500 = 200a

or

a = - 12.5 ft/s²

Here, the negative sign depicts the deceleration.

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The <em>equivalent force-couple system</em> at O is the force and couple experienced when at point O due to the applied force at point A

The <em>equivalent force couple</em> system at O due to force <em>F</em> are;

Force, F =  (<u>8.65·i - 4.6·j</u>) KN

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The reason the above values are correct is as follows:

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The <em>angle </em>away from the vertical the force is applied = 26°

The required parameter:

The <em>equivalent force-couple system</em> at the centroid of the beam cross-section of the cantilever

Solution:

The <em>equivalent force-couple system</em> is the force-couple system that can replace the given force at centroid of the beam cross-section at the cantilever O ;

The <em>equivalent force</em> \overset \longrightarrow F = 9.8 kN × cos(28°)·i - 9.8 kN × sin(28°)·j

Which gives;

The <em>equivalent force</em> \overset \longrightarrow F ≈ (<u>8.65·i - 4.6·j</u>) KN

The <em>couple </em><em>acting </em>at point O due to the force <em>F</em> is given as follows;

The <em>clockwise moment</em> = <em>9.8 kN × cos(28°) × 4.9</em>

The <em>anticlockwise moment</em> = <em>9.8 kN × sin(28°) 0.65/2 </em>

The sum of the moments = Anticlockwise moment - Clockwise moments

∴ The <em>sum </em>of the moments, ∑M, gives the moment acting at point O as follows;

M₀ = <em>9.8 kN × sin(28°) 0.65/2 - 9.8 kN × cos(28°) × 4.9</em>  ≈ 40.9 kN·m

The couple acting at O, due to F,  M₀ ≈ <u>40.9 kN·m</u>

The equivalent force couple system acting at point O due the force, F, is as follows

F =  (8.65·i - 4.6·j) KN

M₀ ≈ <u>40.9 </u>k kN·m

Learn more about equivalent force systems here:

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