A force F acts on a uniform rectangular cabinet weighing 400N as shown in figure. the cabinet slides with constant speed when a force F =200N is applied at height h = 0.4 m at angle $37^{0}$ from horizontal.At what distance x (in cm) from edge A ,the resultant normal reaction acts on cabinet (sin $37^{0}$ = 3/5)

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Solution

The correct option is A 20
$\sum F_{y} = 0 F_{y} + N = W F sin 37 ° + N = 400 200 \times \frac{3}{5} + N = 400 N = 400 - 120 = 280 N \sum M = 0 N_{x} + F_{x} \times 40 = W \times 30$
$280 \times x + 200 \times \frac{4}{5} \times 40 = 400 \times 30 280 x = 12000 - 6400 = 5600 ∴ x = \frac{5600}{28} = 20$

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The correct option is A 20∑Fy=0Fy+N=WFsin37°+N=400200×35+N=400N=400−120=280N∑M=0Nx+Fx×40=W×30280×x+200×45×40=400×30280x=12000−6400=5600∴x=560028=20