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A steel beam that is 6.50 m long weighs 336 N. It rests on two supports, 3.00 m apart, with equal amounts of the beam extending from each end. Suki, who weighs 590 N, stands on the beam in the center and then walks toward one end. How close to the end can she come before the beam begins to tip?

Respuesta :

Answer:

Explanation:

When beam is balanced and not rotating with Suki standing on it , let reaction force on the supports be R₁ and R₂. Then

R₁ +R₂ = 336 + 590

= 929

Now the moment beam begins to tip , reaction on distant support R₁ = 0

only R₂ will exists on the support near to Suki.

Taking torque about this support of weight of beam acting from the middle point and weight of suki of 590N ,who is x distance from the support towards the other end.

336 x 1.5 = 590 x

x = .85 m

ie , from second support , Suki can not go beyond a distance of .85 m towards the second end.

The distance to the end that suki can come before the beam begins to tip is; 0.854 m

We are given;

Length of beam; L = 6.5 m

Weight of beam; W_b = 336 N

Weight of suki; W_s = 590 N

Since the steel beam rests on two supports 3 m apart and suki stands in the beam and walks towards the end, it means that if we take moments about the left support, a distance x from suki to the left will be the distance that suki can come before the beam begins to tip.

Thus;

W_b × (3/2) = (W_s × x)

Plugging in the relevant values gives;

336 × 1.5 = 590 × x

x = (336 × 1.5)/590

x = 0.854 m

Read more about moments about a point on a beam at; https://brainly.com/question/14698103

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