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3 years ago

9

A pair of fuzzy dice is hanging by a string from your rearview mirror. while you are accelerating from a stoplight to 28 m/s in

6.0 s, what angle does the string make with the vertical? (assume your acceleration is constant.)

1 answer:

Lerok [7]3 years ago

4 0

We can use the formula of motion in physics (2nd law od newton) in this problem:
x direction: Fsin ∅ = ma
y direction: Fcos ∅ -mg = 0
∅ is equal to sin ∅ / cos ∅ or x/y
tan ∅ = ma / mg = a /g

Applying acceleration formula:
v = vo + at ; 28 = 0 + 6a ; a = 4.67 m/s^2
∅ = tan-1 (a/g) = tan-1 (4.67/9.81) = <span>25.4 degrees.</span>

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A rectangular key was used in a pulley connected to a line shaft with a power of 7.46 kW at a speed of 1200 rpm. If the shearing

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Speed, N = 1200 rpm

Shearing stress of shaft, $\tau _{shaft}$ = 30 MPa

Shearing stress of key, $\tau _{key}$ = 240 MPa

width of key, w = $\frac{d}{4}$

d is shaft diameter

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Torque, T = $\frac{P}{\omega }$

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$\omega = \frac{2\pi N}{60}$

$T = \frac{7460}{\frac{2\pi (1200 )}{60}}$ = 59.365 N-m

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$\tau _{shaft} = \tau _{max} = \frac{2T}{\pi (\frac{d}{2})^{3}}$

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$\tau _{key}$ = $\frac{F}{wl}$

we know that

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⇒ $\tau _{key}$ = $\frac{\frac{T}{\frac{d}{2}}}{wl}$

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