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Problem 43 Medium Difficulty

A horizontal aluminum rod 4.8 $\mathrm{cm}$ in diameter projects 5.3 $\mathrm{cm}$ from a wall. A 1200 $\mathrm{kg}$ object is suspended from the
end of the rod. The shear modulus of aluminum is $3.0 \times 10^{10} \mathrm{N} / \mathrm{m}^{2}$ .
Neglecting the rod's mass, find (a) the shear stress on the rod and
(b) the vertical deflection of the end of the rod.

Answer

(a) $6.5 \times 10^{6} \mathrm{N} / \mathrm{m}^{2}$
(b) $1.1 \times 10^{-5} \mathrm{m}$

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Video Transcript

for this problem on the topic of equilibrium and elasticity and telemedia aluminum rod that is 4.8 centimeters in diameter is projecting from a wall, 5.3 centimeters. A 1200 kg object is then suspended from the end of the rod. We are given the sheer modular of aluminum and we are asked to calculate the shear stress on the rod as well as the vertical deflection of the end of the rod. Now the weight of the object hung on the end provides the source of this year stress, and the sheer stress is given by F. Of a. Where F. Is the magnitude of this force applied parallel to one of the faces of the aluminum rod and A as its cross sectional area. Now we know that F. Here is equal to the weight of the object M. G. Where M is the mass of the object. The cross sectional area A. Is pie R squared are here is the radius of the road. Now, if we substitute the values given, we can find this year, the shear stress as follows this year stresses F over A. Which is the weight of the object MG. Divided by the circular surface area, pi R squared. And since these values are known, we get this to be 1000 200 kg. The mass of the object times 9.8 m per square second, divided by pie time 0.24 m squared. And so calculating, we get the shear stress to be 6.5 times 10 to the power six newton's per square meter. And so that's the shear stress on the rod. Now, for part B of the problem, we want to find the vertical deflection of the end of the rod. Now, this year modules G. Is given by the force per unit area divided by the extension of the Lord delta X, divided by its original length. L. Now, L. Is simply the protrusion of the rod from the wall. And if we rearrange the situation, we get the extension that require down to X to be F over A. Which we calculated above times our over the sheer module S. G. Which is given for alum in ium. So if we substitute the values into this equation, we get F over A to be 6.5 times 10 to the power six newtons per square meter, multiplied by 0.0 53 m provided by G, which is given as three times 10 to the power 10 newton's per square meter. And so computing this, we get the vertical deflection of the end of the rod X to be 1.1 times 10 to the minus five meters.