1.A system of cables is used to support a crate of mass m = 45 kg as shown in th
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1.A system of cables is used to support a crate of mass m = 45 kg as shown in the figure below (? = 57
A system of cables is used to support a crate of mass m = 45 kg as shown in the figure below (? = 57 degree ). Find the tensions in all three cables. The three forces shown in the figure below act on a particle (with F1 = 46.0 N, F2 = 56.0 N, ?1 = 61.0 degree , and ?2 = 29.0 degree ). If the particle is in translational equilibrium, find F3, the magnitude of force 3 and the angle ?3. Two crates are connected by a massless rope that passes over a pulley as shown in the figure below. If the crates have mass 32 kg and 81 kg, find their acceleration. If the system begins at rest, with the more massive crate a distance 13 m above the floor, how long does it take for the more massive crate to reach the floor? Assume the pulley is massless and frictionless. A swimmer is able to swim at a speed of 0.82 m/s in still water. How long does it take the swimmer to go a distance of 1410 m in still water? Call this time t1. The swimmer then decides to swim the same distance in a river having a current with a speed of 0.54 m/s. How long does this take if she swims with the current? Call this time t2. How long does it take her to swim the same distance against the current? Call this time t3.Explanation / Answer
1)
T2 = mg = 45*9.8 = 441 N
T3*sin(theta) = T2
T3 = 441 / sin(57 deg) = 525.83 N
T1 = T3*cos(theta) = 525.83*cos(57 deg)
T1 = 286.38 N
2)
F3*cos(theta3) = F1*cos(theta1) + F2*cos(theta2)
F3*cos(theta3) = 46*cos(61 deg) + 56*cos(29 deg) = 71.279 N
F3*sin(theta3) = F1*sin(theta1) - F2*sin(theta2)
F3*sin(theta3) = 46*sin(61 deg) - 56*sin(29 deg) = 13.083 N
By solving these two eqns ,
F3 = 72.47 N
theta3 = 10.4 degrees
3)
mb = 32 kg
ma = 81 kg
T = tension
w =weight
sum forces on block A
T = W - F
sum forces on block b
T = W+ F
Ta=Tb
ma*g-ma*a= mb*g + mb*a
a =[(ma-mb)/(ma+mb)]*g = (81-32)*9.8/(81+32) = 4.249 m/s^2
s= distance = 13 m
s = 0.5at^2 +v*t (system begins at rest so v = 0)
t =sqrt(2s/a)= sqrt(2*13/4.249) = 2.473 s
4)
t = d/v
a) t1 = 1410 / 0.82 = 1719.5 s
b) t2 = 1410/(0.54+0.82) = 1036.76 s
c) t3 = 1410 / (0.82-0.54) = 5035.71 s
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