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2 years ago

Define couple and give 2 examples

2 answers:

6 0

A couple consists of two parallel forces that are equal in magnitude, opposite in sense and do not share a line of action. ... For example, the forces that two hands apply to turn a steering wheel are often (or should be) a couple. Each hand grips the wheel at points on opposite sides of the shaft.

Elodia [21]2 years ago

3 0

Answer:

Two equal and opposite parallel forces not acting along the same line, form a couple. A couple is always needed to produce the rotation.

For example, turning a key in a lock and turning a steering wheel.

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A pneumatic system consists of several identical 1.5 inch diameter cylinders to lift 450-pound pallets in a warehouse. the gauge

DerKrebs [107]

The applicable equation:

P = F/A

P = pressure
F = Force or weight
A = surface area

Pressure on each cylinder = (W/n)/A
Where n = number of cylinders. Additionally, pressure in the reservoir is equivalent to the pressure in each cylinder.
Net pressure = 75 - 14.7 = 60.3 psi

Therefore,
60.3 = (W/n)/A = (450/n)/(πD^2/4) = (450/n)/(π*1.5^2/4) = (450/n)/(1.7671)
60.3*1.7671 = 450/n
106.03 = 450/n
n = 450/106.3 = 4.244 ≈ 5

The number of cylinders is 5.

3 0

3 years ago

You are to design a rotating cylindrical axle to lift 800-N buckets of cement from the ground to a rooftop 78.0 m above the grou

Ronch [10]

Answer:

The diameter of the axle is 5.08 cm.

Explanation:

Given that,

Force = 800 N

Distance = 78.0 m

Suppose we need to find the diameter of the axle be in order to raise the buckets at a steady 2.00 cm/s when it is turning at 7.5 rpm.

We need to calculate the radius of axle

Using formula of linear velocity

$v = r\omega$

$r=\dfrac{v}{\omega}$

Where, v =velocity

r = radius

$\omega$ =angular velocity

Put the value into the formula

$r=\dfrac{2.00}{7.5\times\dfrac{2\pi}{60}}$

$r=2.54\ cm$

We need to calculate the diameter of axle

Using formula of diameter

$d=2r$

$d=2\times2.54$

$d=5.08\ cm$

Hence, The diameter of the axle is 5.08 cm.

8 0

3 years ago

A 30.0 kg packing case is initially at rest on the floor of a 1500 kg pickup truck. The coefficient of static friction between t

alexgriva [62]

Answer:

(a). The magnitude of the frictional force acting on the case is 66 N northward.

(b). The magnitude of the friction force acting on the case is 58.8 N southward.

Explanation:

Given that,

Mass of packing case = 30.0 kg

Mass of truck = 1500 kg

Coefficient of static friction = 0.30

Coefficient of kinetic friction = 0.20

Acceleration = 2.20 m/s² northward

(A). We need to calculate the force acting on the case

Using formula of force

$F=ma$

Put the value into the formula

$F=30.0\times2.20$

$F=66\ N$

We need to calculate the maximum static frictional force

Using formula of frictional force

$f_{\mu}=\mu mg$

Put the value into the formula

$F_{\mu}=0.30\times30.0\times9.8$

$F_{\mu}=88.2$

The frictional force is greater then the force acting on the block

Then, The frictional force is 66 N northward.

(B). We need to calculate the force acting on the case

Using formula of force

$F = ma$

Put the value into the formula

$F=30.0\times3.40$

$F=102\ N$

The static friction force is less then the force action on the case

We need to calculate the maximum static frictional force

Using formula of frictional force

$f_{\mu}=\mu mg$

$f_{\mu}=0.20\times30.0\times9.8$

$f_{\mu}=58.8\ N$

So, The magnitude of the friction force acting on the case is 58.8 N southward.

Hence, This is the required solution.

7 0

3 years ago

An infinite line of charge with linear density λ1 = -6.6μC/m is positioned along the axis of a thick conducting shell of inner r

tester [92]

Answer:

1210 N/m between the inner and outer edges (repulsive)

252.59 N/m between inner and line charges (attractive)

1210+252.59=1462.59 N/m effective force/metre on the inner edge

Explanation:

f=k*Qq/r^2 was applied, the charges are concentrated at the edges of a ring both inner and the outer were shape edges with uniform charge densities 6.6 micro C/m of distance 4.5-2.7=1.8 cm

7 0

3 years ago

A 7.80 g bullet has a speed of 620 m/s when it hits a target, causing the target to move 6.30 cm in the direction of the bullet'

Arada [10]

Answer:

$23796.19\ \text{N}$

$0.0002032\ \text{s}$

Explanation:

F = Force

s = Displacement = 6.3 cm

m = Mass of bullet = 7.8 g

v = Velocity of bullet = 620 m/s

t = Time taken

Work done is given by

$W=Fs$

Kinetic energy is given by

$K=\dfrac{1}{2}mv^2$

Using work energy considerations we get

$Fs=\dfrac{1}{2}mv^2\\\Rightarrow F=\dfrac{1}{2s}mv^2\\\Rightarrow F=\dfrac{1}{2\times 0.063}\times 7.8\times 10^{-3}\times 620^2\\\Rightarrow F=23796.19\ \text{N}$

The average force that stops the bullet is $23796.19\ \text{N}$ .

Force is given by

$F=m\dfrac{v-u}{t}\\\Rightarrow t=m\dfrac{v-u}{F}\\\Rightarrow t=7.8\times 10^{-3}\times \dfrac{620}{23796.19}\\\Rightarrow t=0.0002032\ \text{s}$

The time taken to stop the bullet is $0.0002032\ \text{s}$ .

8 0

3 years ago