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2 years ago

Proposed Exercise - Circular Movement

Physics

1 answer:

notka56 [123]2 years ago

5 0

Answer:

ωB = 300 rad/s

ωC = 600 rad/s

Explanation:

The linear velocity of the belt is the same at pulley A as it is at pulley D.

vA = vD

ωA rA = ωD rD

ωD = (rA / rD) ωA

Pulley B has the same angular velocity as pulley D.

ωB = ωD

The linear velocity of the belt is the same at pulley B as it is at pulley C.

vB = vC

ωB rB = ωC rC

ωC = (rB / rC) ωB

Given:

ω₀A = 40 rad/s

αA = 20 rad/s²

t = 3 s

Find: ωA

ω = αt + ω₀

ωA = (20 rad/s²) (3 s) + 40 rad/s

ωA = 100 rad/s

ωD = (rA / rD) ωA = (75 mm / 25 mm) (100 rad/s) = 300 rad/s

ωB = ωD = 300 rad/s

ωC = (rB / rC) ωB = (100 mm / 50 mm) (300 rad/s) = 600 rad/s

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Three crates with various contents are pulled by a force Fpull=3615 N across a horizontal, frictionless roller‑conveyor system.

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The question is incomplete. Here is the complete question.

Three crtaes with various contents are pulled by a force Fpull=3615N across a horizontal, frictionless roller-conveyor system.The group pf boxes accelerates at 1.516m/s2 to the right. Between each adjacent pair of boxes is a force meter that measures the magnitude of the tension in the connecting rope. Between the box of mass m1 and the box of mass m2, the force meter reads F12=1387N. Between the box of mass m2 and box of mass m3, the force meter reads F23=2304N. Assume that the ropes and force meters are massless.

(a) What is the total mass of the three boxes?

(b) What is the mass of each box?

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(b) m1 = 915kg;

m2 = 605kg;

m3 = 864.5kg;

Explanation: The image of the boxes is described in the picture below.

(a) The system is moving at a constant acceleration and with a force Fpull. Using Newton's 2nd Law:

$F_{pull}=m_{T}.a$

$m_{T}=\frac{F_{pull}}{a}$

$m_{T}=\frac{3615}{1.516}$

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(b) For each mass, analyse each box and make them each a free-body diagram.

For $m_{1}$ :

The only force acting On the $m_{1}$ box is force of tension between 1 and 2 and as all the system is moving at a same acceleration.

$m_{1} = \frac{F_{12}}{a}$

$m_{1} = \frac{1387}{1.516}$

$m_{1}$ = 915kg

For $m_{2}$ :

There are two forces acting on $m_{2}$ : tension caused by box 1 and tension caused by box 3. Positive referential is to the right (because it's the movement's direction), so force caused by 1 is opposing force caused by 3:

$m_{2} = \frac{F_{23}-F_{12}}{a}$

$m_{2} = \frac{2304-1387}{1.516}$

$m_{2}$ = 605kg

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$m_{3} = 2384.5-1520.0$

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