Three blocks are arranged in a stack on a frictionless horizontal surface. The bottom block has a mass of 37.0 kg. A block of mass 18.0 kg sits on top of it and a 14.0 kg block sits on top of the middle block. A downward vertical force of 170 N is applied to the top block. What is the magnitude of the normal force exerted by the bottom block on the middle block?
2006-12-08 16:01:08 · 2 answers · asked by Spirit of Vengeance 2 in Science & Mathematics ➔ Physics
F = 170 + (14 + 18)*9.80662
F = 483.81 N
2006-12-08 16:13:26 · answer #1 · answered by Helmut 7 · 0⤊ 0⤋
To have a complete answer start by drawing FBDs for m1 (Mass1=the toppest mass), m2. (No need for m3)
(1) FBD for m1: 1 upward force exerted by m2 to m1(Fn1), 2 downward forces (Fg1+Fapplied)
(2) FBD for m2: 1 upward force (Fn2), 2 downward forces ((Fg1+Fapplied)+Fg2)
Fn2 is what we are looking for! Since we have equilibruim:
Fn2 = Fg1+Fg2+Fapplied = (m1+m2)*g+Fapplied = (18.0kg+14.0kg)*9.81+170N = 483.92N
(I wish I could show it graphically)
2006-12-09 00:59:02 · answer #2 · answered by mh_mn 2 · 0⤊ 0⤋