A rigid rod of mass 0.2 kg and length 1.5 m is suspended by wires at each end. The wireshave equal rest length. The wire on the left end of the rod has Young’s modulus4*10^11Pa, while the one on the right end has Young’s modulus2*10^11Pa. A 2 kg mass is hung fromthe beam. How far from the left end of the rod should the mass be hung such that the bar is horizontal?
I was told I need to use tension some how but I can't seem to get it.
I guess the wires have equal area A
delta L = L * F/(EA) same for each
so
F/E must be the same for each
F1/4 = F2/2
F1 = 2 F2
now moments about distance x from left where m is hung
F1 * x = F2 (1.5-x)
2 F2 x = F2(1.5-x)
3 x = 1.5
x = .5 meters
To solve this problem, you need to consider the equilibrium of forces acting on the rigid rod. Let's break it down step by step:
1. Start by considering the forces acting at each end of the rod. Since the wires have equal rest length, the tension in both wires will be equal when the rod is horizontal.
2. Let's assume the distance from the left end of the rod to where the mass is hung is x. The remaining length of the rod from the mass to the right end will then be (1.5 - x).
3. Now, consider the tension in the wire on the left end of the rod. Since the left wire has a Young's modulus of 4*10^11 Pa, it will stretch by an amount equal to (Tension_left / (A * Young's Modulus_left)), where A is the cross-sectional area of the wire.
4. Similarly, the tension in the wire on the right end of the rod, which has a Young's modulus of 2*10^11 Pa, will stretch by an amount equal to (Tension_right / (A * Young's Modulus_right)).
5. Since the rod is in equilibrium, the stretch in the left wire should be equal to the stretch in the right wire. So, we can set up an equation as: (Tension_left / (A * Young's Modulus_left)) = (Tension_right / (A * Young's Modulus_right)).
6. Solving this equation will give us the relationship between the tensions in the two wires.
7. Now consider the forces acting on the rod. We have the weight of the rod itself, and the weight of the 2 kg mass. These forces act at the center of gravity of each segment of the rod and should balance each other.
8. The weight of the rod can be considered as acting at its center, which is at a distance of 0.75 m from the left end. The weight of the 2 kg mass acts at a distance of (x + (1.5 - x)/2) from the left end.
9. The remaining step is to determine the tensions in the wires. You can consider the torque about the left end of the rod to get the relationship between the weights and the tensions.
10. Equate the relationship obtained in step 6 with the relationship between the weights and the tensions obtained in step 9. Solve this equation to find the value of x that makes the rod horizontal.
This is a basic outline of the steps involved in solving the problem. You will need to apply the relevant equations and solve them to obtain the final value of x.