A block with a mass of 8.7 kg is dropped from rest from a height of 8.7 m, and remains at rest after hitting the ground. 1)If we
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Answer:
Tension in right wire = 25.9N
Explanation:
I have attached a free body diagram to depict this question.
From the diagram, i have labelled the tensions in the strings T1 and T2.
While i labelled the weight of the bar as Wb and weight of sausage as Ws.
Now, when solving a problem like this we want to first remember that the beam is static; meaning it is not moving. From simple physics, this means that the sum of the forces in the y direction equals zero (i.e. the total downward forces equal the total upward forces)
Thus, from the diagram, the upward forces are T1 and T2 while the downward forces are Ws and Wb.
Thus;
T1 + T2 = Wb + Ws
We know that mass of bar = 4.94kg. Thus, Weight of bar(Wb) = mg = 4.94 x 9.81 = 48.46N
Also, weight of sausage (Ws) = mg = 2.49 x 9.81 = 24.43N
Thus,
T1 + T2 = 48.46N + 24.43
T1 + T2 = 72.89N - - - - - (eq 1)
Now, let's take moments about the left end of the bar.
The maximum weight of the bar will act at the centre, so distance from the Wb to left end = 1.46/2 = 0.73m
So, moments about left end;
T2 x 1.46 = (Wb x 0.73) + (Ws x 0.1)
1.46T2 = (48.46 x 0.73) + (24.43 x 0.1)
1.46T2 = 35.373 + 2.443
1.46T2 = 37.816
T2 = 37.816/1.46 = 25.9N
Answer:
A. Kindly find attached free body diagram for your reference (smiles I guess I will make a terrible artist)
B. The collision is inelastic because both the husband and the wife moved together with same velocity as he grabs her on the waist
C. The general equation for conservation of momentum in terms of m 1, v 1, m 2, v 2, and final velocity vf
Say mass of husband is m1
Mass of the wife is m2
Velocity of the husband is v1
Velocity of the wife is v2
According to the conservation of momentum principle momentum before impact m1v1+m2v2 =momentum after impact Common velocity after impact (m1+m2)vf
The momentum equation is
m1v1+m2v2= (m1+m2)vf
D. To solve for vf we need to make it subject of formula
vf= {(m1v1) +(m2v2)}/(m1+m2)
E. Substituting our given data
vf=
{(1570*58)+(2550*54)}/(1570+2558)
vf=91060+137700/4120
vf=228760/4120
vf=55.52m/s
Their speed after collision is 55.52m/s
