A box to be pushed across a smooth floor asshown. The box has a mass of 7.5 kg,
ID: 1744731 • Letter: A
Question
A box to be pushed across a smooth floor asshown.The box has a mass of 7.5 kg, the coefficient of friction betweenthe box and the floor is 0.42, the angle between the direction ofthe force and the horizontal is 31° and the acceleration dueto gravity is 9.8 m.s2. The magnitude of theforce required to just move the box is nearest to:
Student Response Correct Answer Feedback A. 28.8 N B. 90.2 N C. 48.2 N D. 59.9 N E. 30.9 N General Feedback: Sketch the problem. Resolve the force F into verticaland horizontal components. Then sketch these components plus theweight force and the friction force acting on the box.
You are given m = 7.5 kg, = 0.42, = 31° and g = 9.8m.s2.
The nett vertical force downwards (the Normal force) is N= mg + Fsin
The nett horizontal force which will just cause slipping occurswhen
Fcos = N = (mg + Fsin)
make F the subject of this expression to calculate the answer.
Explanation / Answer
Given : m= 7.5 kg, = 0.42, = 31° andg = 9.8 m.s2.The nett vertical force downwards (the Normal force) isN = mg + Fsin
The nett horizontal force which will just cause slippingoccurs when
Fcos = N = ( mg +Fsin) or Fcos - F sin = m g F ( cos - sin ) = m g or F = ( m g) / ( cos - sin ) = ( 0.42*7.5*9.8) / (cos31 - 0.42* sin31) = 30.87 / 0.64084 = 48.17 ˜ 48.2 N Hope this helps u!
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