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3 years ago

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How do you find the parallel component of force (Fsub||) on an inclined plane? Is it equal to the y-component? I also need to fi

nd applied force.
Mass = 100 kg
Weight = 980 N
Normal Force = 965 N
Theta = 10°
Friction = 289.5 N

Please Help :)

1 answer:

stepladder [879]3 years ago

6 0

Answer:

just add them all up i think i dont know

Explanation:

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Explanation:

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3 years ago

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Answer:

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$T = 342.5 \hat i + 675\hat j$

Part b)

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Part c)

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Part d)

$\theta = 333 degree$

Part e)

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Part f)

$\theta = 333 degree$

Explanation:

Part a)

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$T = 757 N$ at 26.9 degree with vertical

$T = 757 sin26.9 \hat i + 757 cos26.9 \hat j$

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Part b)

Net force on Tarzen is given as

$F_{net} = T + F_g$

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Part c)

magnitude of the force is given as

$F = \sqrt{F_x^2 + F_y^2}$

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$F = 384.2 N$

Part d)

Direction of the force is given as

$tan\theta = \frac{F_y}{F_x}$

$tan\theta = \frac{-174}{342.5}$

$\theta = 333 degree$

Part e)

Magnitude of the acceleration

$a = \frac{F}{m}$

$m = \frac{849}{9.81} = 86.5 kg$

tex]a = \frac{384.2}{86.5}[/tex]

$a = 4.4 m/s^2$

Part f)

Direction of acceleration is same as the direction of the force

$\theta = 333 degree$

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3 years ago