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4 years ago

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A uniform plank 6.1 m long rests on two supports, 2.5 m apart. The gravitational force on the plank is 127 N. The left end of th

e plank is 1.5 m to the left of the left support, so the plank is not centered on the supports. A person is standing on the plank one tenth of a meter to the right of the right support. The gravitational force on this person is 846 N. How far to the right can the person walk before the plank begins to tip

1 answer:

Digiron [165]4 years ago

6 0

Answer:

The person can walk 0.0426 m to the right

Explanation:

Moment = Force * Perpendicular moment

The principle of moment states that the sum of clockwise moments should equal the sum of the anticlockwise moments

From the free body diagram attached to this solution. If the moment is taken about B

The length of the plank = 6.1 m

The weight of the plank will act at the middle i.e. 6.1/2 = 3.05 m

Sum of clockwise moments = Sum of anticlockwise moments

(127 * 1.5) + (127*2.5) = (127 * 3.05) + (846 * l)

190.5 + 317.5 = 387.35 + 846l

190.5 + 317.5 - 387.35 = 846l

120.65 = 846l

l = 120.65/846

l = 0.1426 m

Since the person is standing at the 0.1 m to the right of the support, that means he can only walk a distance of 0.1426 m - 0.1 m to the right

The distance the person can walk to the right = 0.0426 m to the right

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$▪▪▪▪▪▪▪▪▪▪▪▪▪ {\huge\mathfrak{Answer}}▪▪▪▪▪▪▪▪▪▪▪▪▪▪$

Distance travelled by the truck is ~

$\boxed{240 \: \: km}$

And it's displacement is ~

$\boxed{0 \: \: km}$

$\large \boxed{ \mathfrak{Step\:\: By\:\:Step\:\:Explanation}}$

See the diagram in attachment for reference ~

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