Physics problem involving friction help
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Physics problem involving friction help

[From: ] [author: ] [Date: 12-06-21] [Hit: ]
F - (0.75)(30) - 0.5(30 + 100) = (100/g)(0.F = 162.......
A small box weigh 30N is on top of a bigger box weigh 100N. The coefficient of kinetic friction between block and the floor is 0.5. The coefficient of static friction between the two blocks is 0.75. What is the maximum horizontal force that can be exerted on the lower block before the top block begins to slide off?

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freebody of upper block,

ΣF = m a

μ N = (N/g) a

0.75 * 30 = (30/g) a

a = 0.75 * 30/(30/g). . . . . . (equation 1)

freebody of lower block,

ΣF = m a

F - μ N - μ' (N + N') = (N'/g) a

F - (0.75)(30) - 0.5(30 + 100) = (100/g) a . . . . . . (equation 2)

substitute eqn 1 to eqn 2,

F - (0.75)(30) - 0.5(30 + 100) = (100/g)(0.75 * 30/(30/g))

F = 162.5 Newton
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