Horizontal Sausage
Horizontal Sausage
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Please help with physics problem!!!?
In a butcher shop, a horizontal steel bar of mass 4.04 kg and length 1.20 m is supported by two vertical wires attached to its ends. The butcher hangs a sausage of mass 2.61 kg from a hook that is at a distance of 0.23 m from the left end of the bar. What are the tensions in the two wires?
I'm not getting the right answer for that. Plus, I thought tension was suppose to be measured in Newtons.
In a butcher shop, a horizontal steel bar of mass 4.04 kg and length 1.20 m is supported by two vertical wires attached to its ends. The butcher hangs a sausage of mass 2.61 kg from a hook that is at a distance of 0.23 m from the left end of the bar. What are the tensions in the two wires?
The center of mass of the 1.20 m long steel bar is located at the center, which is 0.60 m from the left end. The 2.61 kg sausage is 0.23 m from the left end.
Since the sausage is closer to the left end than the right end, the left wire must support more of the total weight than the right wire!
Weight of bar = 4.04 * 9.8 = 39.592 N
Weight of sausage = 2.61 * 9.8 = 25.578 N
The sum of the tensions in the two wires must support the total weight!
Tl + Tr = 39.592 + 25.578 = 65.17 N
Eq. #1 Tl + Tr = 65.17 N
The clockwise torque must equal the counter clockwise torque!
I choose the pivot point at the left end.
clockwise torque = (39.592 * 0.60) + (25.578 * 0.23)
counter clockwise torque = Tr * 1.20
Eq. #2
Tr * 1.20 = (39.592 * 0.60) + (25.578 * 0.23) = 29.63814
Solve for Tr, then solve for Tl in Eq. #1
Tr * 1.20 = 29.63814
Tright = 24.69845 Newtons
Tl + 24.69845 = 65.17 N
Tleft = 40.47155 Newtons
