Question

Principal . Anyw a Weight plank weight block Equibrium of a Plank under 4 forces (Two Choices of Rotational Axis) As shown in the figure, a uniform plank of length - 4.4 m rests on two supports that are D - 2.7 m apart. The mass of the plank is meant - 28.0 kg and its center of mass is at the midpoint. A 4.0 kg block is placed on the right end of the plank. Find the magnitude of the two normal forces N, and Ny Keep 2 decimal places. Apply the first condition of equilibrium: Net force - 0. Obtain first equation involing N1 and N2. You will try two choices of rotational axis for torque calculations. Choice 1: the LEFT support as rotational axis. It is good practice to mark the rotational axis on the drawing. From this rotational axis, it is straightforward to figure out the lever arm for each force. Write each torque, pay attention to the direction of each torque. Apply the second condition of equilibrium: Net torque - 0. Find the magnitude in Newtons of force Ny N Submit 15 attempts remaining) Then Find the Magnitude (in Newtons) of N, ter number N

Answer 1

V the my Weight of block Wp-mpg ID - L: 4,47 10=2-7m , mp= 28.01g Макс Вык, тч.bg For equilibrium (1) Efy=0 → Ni+N2 = mpg + mg Ni+N2 = (marmolg = (28+4)*9*8 N = 313•6N . Ni+ N2=313•6N (2) Net for que is zero claro f Takke votational axis at point al > - mpg [ + N2D - mogu =0. → N2 = (mp + 2 mb)gL - 20 N₂ = (28+2x4) x 9.8x 4.4 N - 2x 2.7 [12=287.50 from equation @ Ni+N2=313-6N 3 Ni=1313.6-287-5 )N N1=261 N we cansider rotational axis at 'B' then for equilibrium ETD=0 5 -Nyx + feruj mpegi - mbg) (L-) =0 ² x (-Ni+mpg + mog) = Impg + Lm bg 3txo za LC mp+2 mi) 2 (mog+ming -N1 xa 414x9.84 (28+2x4) m 2.28+4)3.8 * - 26.1] = 206997 m ~ 2-77 - x = 2.70 + lever arm for force Ni For weight of the plance lever arm , p = (x-t1 . X=(2.7-44) os 20.5m. .. slp=0.5m) lever arm for the weight of block alb= L-X ci 34= (4.4-2.7) = 4= 1.7m)