While moving in, a new homeowner is pushing a box across the floor at a constant velocity. the coefficient of kinetic friction between the box and the floor is 0.382. the pushing force is directed downward at an angle θ below the horizontal. when θ is greater than a certain value, it is not possible to move the box, no matter how large the pushing force is. find that value of θ.

θ = 69°

Explanation:

With a sum of forces on the box:

x-axis: F*cosθ - Ff = 0

y-axis: N - m*g - F*sinθ = 0    From this equation we find N:

N = m*g + F*sinθ

Knowing that Ff = μ*N:

F*cosθ - μ*(m*g + F*sinθ) = 0

F*(cosθ - μ*sinθ) = μ*m*g

(cosθ - μ*sinθ) = μ*m*g / F   As you can see here, when F is a lot greater than μ*m*g, the 2nd member of the equation is almost 0:

(cosθ - μ*sinθ) = 0

Solving for θ:

θ = atan( 1/μ ) = 69°   With any angle less than this, the box could still be moved.

acceleration = (change in speed) / (time for the change)

change in speed = (20-120)= -100 m/s.

1.5 min = 90 sec

acceleration = -100/90 = -1.11 m/s^2.

we don't care about the 150kg or the 2,500 meters. they don't make any difference.

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the answer is f1 and d2 because if you put equal force on those two points the beam would balance trust me on this