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120N
30kg
20kg
50kg
120N
F = F - F
Both boxes have friction with surface at μ = 0.2
net
app
fric
F = 120 - 50(9.8)(0.2)
net
F = 120 - 98 = 22N
net
22 = ma 22 = 50(a)
a = 0.44m/s^2
F = F - F - F
net
app
contact
fric
13.2 = 120 - F - 30(9.8)(0.2)
contact
F = ma = 30(0.44) = 13.2N
net
F = -48N
contact
this is the force that both the 30kg and the 20kg extert on eachother
20kg
48N
F = F - F - F
net
app
contact
fric
8.8 = 48 - F - 20(9.8)(0.2)
contact
F = ma = 20(0.44) = 8.8N
net
F = -48N
contact
this is the force that both the 30kg and the 20kg extert on eachother