A 110 kg lower half of a doublehung window has two ropes at

A 11.0 kg lower half of a double-hung window has two ropes at the sides, each rope passing upward over a pulley and each supporting a 5.5kg counterweight. The counterweights are free to move, the pulleys may be considered massless and frictionless, and the window merely rests against the frame. The coefficient of friction between the window and frame is 0.300. In raising it with a pole, a force is applied at the middle of the window of the window in an upward direction at an angle of 15deg to the vertical. How much force must be applied in this manner to raise the window 1.00 m in 2.00 seconds starting from rest?

Solution

Acceleration is calculated as follows

s=[1/2]at^2

1 = [1/2]×a×2^2

Acceleration is, a=0.5 m/s^2

Hence the force on the window:

F= mg sin@ + mg-umgcos@

=107.97 N