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Simora [160]
3 years ago
12

Determine the velocity of the 60-lb block A if the two blocks are released from rest and the 40-lb block B moves 2 ft up the inc

line. The coefficient of kinetic friction between both blocks and the inclined planes is μk=0.10.

Engineering
1 answer:
Otrada [13]3 years ago
5 0

vₐ = 0.771 ft/s

vb = -1.54 ft/s

<u />

<u>Explanation:</u>

<u />

Block A:

F = ma

Nₐ = - 60 cos60° = 0

Nₐ = 30 lb

Fₐ = 0.1 X 30 = 3lb

Block B:

F = ma

Nb = - 40 cos30° = 0

Nb = 34.64lb

Fb = 0.1 X 34.64 = 3.464lb

T1 + ∑U = T2

(0 + 0) + 60 sin 60° |Δsₐ| - 40 sin 30° |Δsb| - 3|Δsₐ| - 3.464 |Δsb|

= 1/2 (60/32.2) vₐ² + 1/2 (40/32.2) vb²

and

2vₐ = - vb

On solving, we get

vₐ = 0.771 ft/s

vb = -1.54 ft/s

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Explanation:

Given;

number of turns, N = 179

radius of the circular coil, r = 3.95 cm = 0.0395 m

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