Electrical pressure or “force” A) current B) resistance C) voltage
1 answer:
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Answer:
For block A, a = 9.66 ft/s²
For block B, a = 15 ft/s²
Explanation:
A free body diagram for this force system is attached to this solution
Mass of block A = m₁ = 8 lb
Mass of block B = m₂ = 6 lb
Coefficient of kinetic friction = μ
Normal reaction on the blocks = N
Spring stiffness of the spring btw block A and B = k = 20 lb/ft
Compression of the spring = 0.2 ft
Analysing Block A first
The forces on block A include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 8 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 8/32.2 = 0.248 lbm
(20×0.2) - (0.2 × 8) = (0.248) aₓ
aₓ = 9.66 ft/s²
The forces on block B include, the weight, normal reaction, frictional force and the elastic force due to the spring
Sum of forces in the y-direction = 0
So, the weight of the block = Normal reaction of the surface on the block
N = W = 6 lb
Sum of forces in the x-direction = maₓ
(k × x) - (μ × N) = maₓ
m = W/g = 6/32.2 = 0.186 lbm
(20×0.2) - (0.2 × 6) = (0.186) aₓ
aₓ = 15 ft/s²
Answer:
R = 4.75 lb (↑)
Explanation:
Number of books = n = 19
Weight of each book = W = 1 lb
Length of the bookshelf = L = 40 inches
We can get the value of the distributed load as follows
q = n*W/L = 19*1 lb/ 40 inches = 0.475 lb/in
then the reactions at 4 ends (supports) of the bookshelf are
R = (q/2)/2 = 4.75 lb
We can see the bookshelf in the pic.
