bus weighing 500 kg on a slope that makes an angle 60 degrees with the horizontal the component of bus weight parallel to the slope is (A) 3500sqrt(3) * N (B) 1500sqrt(3) * N (C) 2500sqrt(3) * N (D) 2500N
Asked by unknownpass66 | 4th Jul, 2021, 11:28: AM
if m is mass of the bus , then its weight mg is acting vertically.
If the slope of the inclined surface is 60o , then component of weight parallel to inclined surgface is ( mg sin30 )
as shown in figure .
where g is acceleration due to gravity and it is assumed as g = 10 m/s2
Hence component of weight parallel to inclined surface = mg sin30 = 500 × 10 × sin30 = 2500 N
Answered by Thiyagarajan K | 4th Jul, 2021, 01:30: PM
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