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2 years ago

Please help me on calculating Net Force, examples E and D in the attachment. Would be grateful if you'd explain. Thank you.

Physics

1 answer:

Yanka [14]2 years ago

8 0

The net force is force applied + force friction ( which is negative ) Unfortunately for some reason the numbers are negative, could you ask your teach about that?

Btw, left is usually force fric. which makes it negative

(-750+-250+-550)+1563 = net force

(-792+-632+-100)+(400+900) = net force

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Answer:

Part a)

$F = 135.7 N$

Part b)

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Explanation:

Part a)

If block is sliding up then net force must be zero and friction will be in opposite to the direction of motion of the block

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so we have

$Fcos\theta = mg + \mu(Fsin\theta)$

$F(cos\theta - \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta - \mu sin\theta}$

$F = \frac{55}{cos50 - 0.310(sin50)}$

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Part b)

If block is sliding down then net force must be zero and friction will be in opposite to the direction of motion of the block

$Fcos\theta = mg - F_f$

$Fsin\theta = F_n$

so we have

$Fcos\theta = mg - \mu(Fsin\theta)$

$F(cos\theta + \mu sin\theta) = mg$

$F = \frac{mg}{cos\theta + \mu sin\theta}$

$F = \frac{55}{cos50 + 0.310(sin50)}$

$F = 62.5 N$

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3 years ago

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