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Batfink, who has a mass of 75 kg isplaced in a 25 kg stationary barrel.What is the Fg on Batfink and thebarrel?
SOLUTION:Force of gravity on Batfink and the barrel.
mbf = 75 kg Fg
mb = 25 kgg = -9.8 m/s2
Fg= -980 N
Batfink and the barrel are raised at1.1 m/s2. What is the force ofsupport acting on Batfink and theBarrel?
SOLUTION:Force of support on Batfink and the barrel.
mbf = 75 kg Fs
mb = 25 kgg = -9.8 m/s2
a = 1.1 m/s2
Fs= 1090 N
Suddenly, Batfink and thebarrel are lowered at 1.1m/s2. What is the force ofsupport acting on Batfinkand the Barrel?
SOLUTION:Force of support on Batfink and the barrel.
mbf = 75 kg Fs
mb = 25 kgg = -9.8 m/s2
a = -1.1 m/s2
Fs= 870 N
Hugo Ago-go pushes the barrel with Batfink in it towards the end of the cliff with a 100 N force over a distance of 10 m before the barrel leaves the cliff. The force of friction is 8 N. Draw a force diagram of the situation.
y
Fs 980 N
Ff - 8 N Fa 100 N
x Fg -980 N
Calculate the acceleration
of the barrel in the xaxis.
SOLUTION:Acceleration of Batfink and the barrel.
mbf = 75 kg amb = 25 kgg = -9.8 m/s2
Fa = 100 NFf = 8 NXi = 0 mXf = 10 m a = 0.92 m/s2
What is the Vf of thebarrel just before it fallsoff the cliff?
SOLUTION:Final velocity of Batfink and the barrel.
Vi = 0 m/s Vf
a = 0.92 m/s2 tf
Xi = 0 mXf = 10 mTi = 0 s V = 4.29 m/s
The Incredible Hulk is hanging motionless off the ground by chains attached separately to his wrists from two different walls. The Hulk has a mass of 355 kg. The chain on his right wrist (T1) forms an angle of 26˚ relative to the floor, and the chain from his left wrist (T2) forms an angle of 32˚ relative to the floor. T2 has 2500 N acting on it. Draw a force diagram of the situation.
y
T1 T2 = 2500 N
26° 32°
Fg -3479 N
Determine the tension in T2X.
SOLUTION:Tension in chain #1. First find T2x.
T2 = 2500 N T2x
θ = 32°m = 355 kgg = -9.8 m/s2 Fg = -3479 N T2x = 2120 N
SOLUTION:Tension in chain #1.
T2 = 2500 N T1
θ = 32°m = 355 kgg = -9.8 m/s2 Fg = -3479 NT2x = 2120 N T1 = 2358.2 N
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