A deli owner creates a lunch special display board by taking a uniform board which has a weight of 217 n, cutting it in half, hi
nging the halves together with a frictionless hinge, and setting it up as an inverted "v". determine the minimum coefficient of static friction needed between the board and the ground in order for her to set the display board up with an angle of 30° between the two sides
1 answer:
Refer to the figure shown below.
W = 217/2 = 108.5 N, the weight of one half of the board.
N = W = 108.5 N, the normal reaction at B or C.
R = frictional force at B or C preventing the board from sliding.
The vertical dashed line through A is a line of symmetry.
By definition,
R = μN = 108.5μ N
where
μ = the static coefficient of friction between the board and the ground.
From geometry,
h = 2a tan(30°) = 1.1547a
Take moments about A for the member AB.
2aN - Rh -Wa = 0
2a(108.5) - 108.5μ(1.1547a) - 108.5 a = 0
217 - 125.285μ - 108.5 = 0
125.285μ = 108.5
μ = 0.866
This is the minimum required static coefficient of friction
Answer: 0.866
W = 217/2 = 108.5 N, the weight of one half of the board.
N = W = 108.5 N, the normal reaction at B or C.
R = frictional force at B or C preventing the board from sliding.
The vertical dashed line through A is a line of symmetry.
By definition,
R = μN = 108.5μ N
where
μ = the static coefficient of friction between the board and the ground.
From geometry,
h = 2a tan(30°) = 1.1547a
Take moments about A for the member AB.
2aN - Rh -Wa = 0
2a(108.5) - 108.5μ(1.1547a) - 108.5 a = 0
217 - 125.285μ - 108.5 = 0
125.285μ = 108.5
μ = 0.866
This is the minimum required static coefficient of friction
Answer: 0.866
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