Answered

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A 5.0-kg box is pulled by a horizontal force F applied to the top of the box. When the box meets a low doorstep, it begins to rotate around it. The box is 60 cmcm wide and 70 cmcm high. What minimum magnitude of the force F is needed to cause this movement

Sagot :

nuhulawal20

Answer:

the required minimum magnitude of the force F is 21 N

Explanation:

Given the data in the question,

m = 5 kg

width  = 60 cm

height = 80 cm

Let force is F represent in the image below,

so when the block about to rotate normal shifted to edge of cube

mg(w/2) = Fh

F = mg(w/2) / h

we know that g = 9.8 m/s²

we substitute

F = (5 × 9.8 ( 60/2)) / 70

F = (5 × 9.8 × 30 ) / 70

F = 1470 / 70

F = 21 N

Therefore, the required minimum magnitude of the force F is 21 N

View image nuhulawal20
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