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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 the 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

Respuesta :

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

Ver imagen kollybaba55

The person should be walk 0.0426 m to the right hand side.

Calculation of the distance:

Since we know that

Moment = Force * Perpendicular moment

Also,

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

So we can say that

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

Learn more about force here: https://brainly.com/question/21422036

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