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3 years ago

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A 1 490-kg automobile has a wheel base (the distance between the axles) of 3.10 m. The automobile's center of mass is on the cen

terline at a point 1.10 m behind the front axle. Find the force exerted by the ground on each wheel.

1 answer:

Kobotan [32]3 years ago

4 0

Answer:

Force on each front wheel = 4711 N

Force on each back wheel = 2591 N

Explanation:

Weight of car = 1490 Kg

Distance of COG from front axle= 1.1 m

Distance of COG from back axle = 3.1 - 1.1 = 2 m

Vertical forces are balanced ,

So, R1 + R2 = mg

R1 + R2 = 1490× 9.8

Thus; R1 + R2 = 14602 - - - eq1

Where

R1 is force on front wheel

R2 is force on back wheel

Now, we know that;

Angular momentum about centre of gravity is zero.

Therefore,

R1 × 1.1 = R2 × 2

R1 = (2/1.1)R2

Put this into equation1, to get;

(2/1.1)R2 + R2 = 14602

R2(2.818) = 14602

R2 = 14602/2.818

R2 = 5182 N and R1 = (2/1.1)5182 = 9422 N

Hence, force on each back wheel = R2/2 = 5182/2 = 2591 N

Force on each front wheel = R1/2 = 9422/2 = 4711 N

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A compact car has a mass of 1380 kg . Assume that the car has one spring on each wheel, that the springs are identical, and that

astraxan [27]

Answer:

A) $k=34867.3384\ N.m^{-1}$

B) $\omega'\approx84\ Hz$

Explanation:

Given:

mass of car, $m=1380\ kg$

A)

frequency of spring oscillation, $f=1.6\ Hz$

We knkow the formula for spring oscillation frequency:

$\omega=2\pi.f$

$\Rightarrow \sqrt{\frac{k_{eq}}{m} } =2\pi.f$

$\sqrt{\frac{k_{eq}}{1380} } =2\times \pi\times 1.6$

$k_{eq}=139469.3537\ N.m^{-1}$

Now as we know that the springs are in parallel and their stiffness constant gets added up in parallel.

<u>So, the stiffness of each spring is (as they are identical):</u>

$k=\frac{k_{eq}}{4}$

$k=\frac{139469.3537}{4}$

$k=34867.3384\ N.m^{-1}$

B)

given that 4 passengers of mass 70 kg each are in the car, then the oscillation frequency:

$\omega'=\sqrt{\frac{k_{eq}}{(m+70\times 4)} }$

$\omega'=\sqrt{\frac{139469.3537}{(1380+280)} }$

$\omega'\approx84\ Hz$

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4 years ago

The flywheel of a steam engine runs with a constant angular speed of 113 $rev/min$. When steam is shut off, the friction of the

Oksana_A [137]

Answer:

α = - 1.883 rev/min²

Explanation:

Given

ωin = 113 rev/min

ωfin = 0 rev/min

t = 1.0 h = 60 min

α = ?

we can use the following equation

ωfin = ωin + α*t ⇒ α = (ωfin - ωin) / t

⇒ α = (0 rev/min - 113 rev/min) / (60 min)

⇒ α = - 1.883 rev/min²

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3 years ago

A 30.0-kg girl in a swing is pushed to one side and held at rest by a horizontal force \vec{F} F ⃗ so that the swing ropes

Virty [35]

Answer:

169.74 N

Explanation:

Given,

Mass of the girl = 30 Kg

angle of the rope with vertical, θ = 30°

equating the vertical component of the tension

vertical component of the tension is equal to the weight of the girl.

T cos θ = m g

T cos 30° = 30 x 9.8

T = 339.48 N

Tension on the two ropes is equal to 339.48 N

Tension in each of the rope = T/2

= 339.48/2 = 169.74 N

Hence, the tension in each of the rope is equal to 169.74 N

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3 years ago

Estimate the wavelength of electrons that have been accelerated from rest through a potential difference of 60 kV.

ivolga24 [154]

Answer: 2.068* $10^{-14}$ m

Explanation: According to work energy-theorem , the workdone in accelerating the electron equals the energy it would give off in terms of light.

workdone= qV

energy = hc/λ

q=magnitude of an electronic charge= 1.602* $10^{-16}$

h= planck constant = 6.626* $10^{-34}$

c= speed of light =2.998* $10^{8}$

v= potential difference= 6* $10^{4}$

λ= wavelength=unknown

by making λ subject of formulae we have that

λ= $\frac{hc}{qv}$

λ = 6.626* $10^{-34}$ * 2.998* $10^{8}$ / 1.602* $10^{-16}$ * 6* $10^{4}$

λ = $\frac{19.878*10^{-26} }{9.612*10^{-12} }$

by doing the necessary calculations, we have that

λ = 2.068* $10^{-14}$ m

8 0

3 years ago

a rocket, initially at rest, is fired vertically with a net upward acceleration of 12 m/s2 . at an altitude of 0.50 km, the engi

kobusy [5.1K]

The rocket travelled a maximum height at 1.0102 km.

Given,

The acceleration of a rocket (a) = 12 m/s²

The altitude of the rocket (s) = 0.50 km = 0.5×10³m

The maximum height of the rocket (h) = ?

Solution,

A rocket is a spacecraft, aircraft, vehicle or projectile that obtains thrust from a rocket engine.

The rate of change of the velocity of an object with respect to time is known as acceleration. It is denoted by (a).i.e. unit is m/s²

(a) = Δv/Δt

Where , Δv is change in velocity and Δt is change in time.

The rate of change in position with respect to time is known as velocity. i.e. Its unit is m/s.

(v)= Δx/Δt

Where,Δx is the change in position and Δt is change in time & v is velocity.

Therefore we know the equation of motion is written as,

v² = u² +2as

Where, v is final velocity , u is initial velocity , a is acceleration and s is altitude of the rocket.

Then putting the value ,

v² = 0 + ( 2× 10 × 0.5×10³)m/s

v² = $\sqrt{10000}$ m/s

v = 100 m/s

Therefore, at altitude of 0.50 km the initial velocity of rocket (u) will be 100 m/s, final velocity v become zero and under free falling the acceleration will be taken (-g) then equation of motion can be given as ,

v² = u² - 2(g)h

h = (v²- u² ) / 2g

h = 10,000/2×9.8

h = 510.2 m

So that the rocket travelled the maximum height ,

(h)= (0.5 km + 510.2m)

(h) = 1.0102 km

Hence, the rocket travelled at the maximum height h is 1.0102 km

To know more about acceleration

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4 0

2 years ago