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nataly862011 [7]
2 years ago
8

At a given instant the bottom A of the ladder has an acceleration aA = 4 f t/s2 and velocity vA = 6 f t/s, both acting to the le

ft. Determine the acceleration of the top of the ladder, B, and the ladder’s angular acceleration at this same instant.
Physics
1 answer:
Nana76 [90]2 years ago
6 0

Answer:

Acceleration=24.9ft^2/s^2

Angular acceleration=1.47rads/s

Explanation:

Note before the ladder is inclined at 30° to the horizontal with a length of 16ft

Hence angular velocity = 6/8=0.75rad/s

acceleration Ab=Aa +(Ab/a)+(Ab/a)t

4+0.75^2*16+a*16

0=0.75^2*16cos30°-a*16sin30°---1

Ab=0+0.75^2sin30°+a*16cos30°----2

Solving equation 1

(0.75^2*16cos30/16sin30)=angular acceleration=a=1.47rad/s

Also from equation 2

Ab=0.75^2*16sin30+1.47*16cos30=24.9ft^2/s^2

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Part of the table of values is as follows;

t,                                     x,                                          v

0,                                    0,                                        0

0.4,                                 0.095663,                         0.478316

0.8,                                 0.382653,                          0.956632

1.2,                                   0.860969,                         1.434948

1.6,                                   1.530611,                            1.913264

2,                                     2.39158,                              2.39158

2.4,                                  3.443875,                           2.869896

2.8,                                   4.687497,                          3.348212

3.2,                                   6.122445,                           3.826528

3.6,                                   7.748719,                             4.304844

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