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
