Ahorizontal aluminum rod 3.3 cm in diameter projects 7.0 cm from a wall. a 1000 kg object is suspended from the end of the rod. the shear modulus of aluminum is 3.0·1010 n/m2. neglecting the rod's mass, find (a) the shear stress on the rod and (b) the vertical deflection of the end of the rod.

a) τ= 2.59 MPa

b)y=28.4 x 10⁻⁵ cm

Explanation:

Given that

L= 3.3 cm

d= 7 cm

m= 1000 kg

G= 3 x 10¹⁰ Pa

Force due to load 1000 kg

P = m g= 1000 x 10 N

P = 10000 N

Area A

A=3848.45 mm²

The shear stress τ

τ= 2.59 MPa

Lets take vertical deflection is y

We know that

τ = G Φ

For small deflection

So

τ L = G y

y=0.000284 cm

y=28.4 x 10⁻⁵ cm

step-by-step explanation:

we are given

we have to simplify it

firstly, we will factor it

now, we can cancel like terms

now, we can combine it

so, we get

real, smaller and upside-down.   so the answer should be b.