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The diagram shows a 15-kg box resting on a wedge which has an angle of inclination of 30° and a coefficient of friction of 0.15. Find the value of the vector Fg.

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We calculate first for the normal force, Fn, which is equal to the product of the force and the cosine of the given angle,
                                         Fn = (15 kg)(9.8 m/s²)(cos 30°)
                                          Fn = 127.30 N
Then, to calculate for the Fg, we multiply the normal force by the given coefficient.
                                         Fg = (127.30 N)(0.15) = 19 N
The nearest value is letter A. 
The correct answer is B. 147.15 N

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