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.
1 answer:
vₐ = 0.771 ft/s
vb = -1.54 ft/s
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<u>Explanation:</u>
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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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Answer:
M_AD = 1359.17 lb-in
Explanation:
Given:
- T_ef = 46 lb
Find:
- Moment of that force T_ef about the line joining points A and D.
Solution:
- Find the position of point E:
mag(BC) = sqrt ( 48^2 + 36^2) = 60 in
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Hence, E = < 0.75*48 , 96 , 36*0.75> = < 36 , 96 , 27 > in
- Find unit vector EF:
mag(EF) = sqrt ( (21-36)^2 + (96+14)^2 + (57-27)^2 ) = 115 in
vec(EF) = < -15 , -110 , 30 >
unit(EF) = (1/115) * < -15 , -110 , 30 >
- Tension T_EF = (46/115) * < -15 , -110 , 30 > = < -6 , -44 , 12 > lb
- Find unit vector AD:
mag(AD) = sqrt ( (48)^2 + (-12)^2 + (36)^2 ) = 12*sqrt(26) in
vec(AD) = < 48 , -12 , 36 >
unit(AD) = (1/12*sqrt(26)) * < 48 , -12 , 36 >
unit (AD) = <0.7845 , -0.19612 , 0.58835 >
Next:
M_AD = unit(AD) . ( E x T_EF)
M_AD = 1835.73 + 116.49528 - 593.0568
M_AD = 1359.17 lb-in
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