Place these events in chronological order. Put the numbers 1-6 next to the<br> events.

A slender rod AB, of weight W, is attached to blocks A and B, which move freely in the guides shown. The blocks are connected by

gregori [183]

Answer:

(a) T = W/2(1-tanθ) (b) 39.81°

Explanation:

(a) The equation for tension (T) can be derived by considering the summation of moment in the clockwise direction. Thus:

Summation of moment in clockwise direction is equivalent to zero. Therefore,

T*l*(sinθ) + W*(l/2)*cosθ - T*l*cosθ = 0

T*l*(cosθ - sinθ) = W*(l/2)*cosθ

T = W*cosθ/2(cosθ - sinθ)

Dividing both the numerator and denominator by cosθ, we have:

T = [W*cosθ/cosθ]/2[(cosθ - sinθ)/cosθ] = W/2(1-tanθ)

(b) If T = 3W, then:

3W = W/2(1-tanθ),

Further simplification and rearrangement lead to:

1 - tanθ = 1/6

tanθ = 1 - (1/6) = 5/6

θ = tan^(-1) 5/6 = 39.81°

8 0

3 years ago

A fluid has a dynamic viscosity of 0.048 Pa.s and a specific gravity of 0.913. For the flow of such a fluid over a flat solid su

sattari [20]

Answer:

Explanation:

First we should recall how Newton's laws relates shear stress to a fluid's velocity profile:

$\tau = \mu \cfrac{\partial v}{\partial y}$

where tau is the shear stress, mu is viscosity, v is the fluid's velocity and y is the direction perpendicular to flow.

Now, in this case we have a parabolic velocity profile, and also we know that the fluid's velocity is zero at the boundary (no-slip condition) and that the vertex (maximum) is at $y=75 \, mm$ and the velocity at that point is $1.125 \, m/s$

We can put that in mathematical terms as:

$v(y)= A+ By +Cy^2 \\v(0) = 0\\v(75 \, mm) = 1.125 \, m/s\\v'(75 \, mm) = 0\\$

From the no-slip condition, we can deduce that $A=0$ and so we are left with just two terms:

$v(y) = By + C y ^2 \\$

We know that the vertex is at $y= 75 \, mm$ and so we can rewrite the last equation as:

$v(y) = k(y-75 \, mm) ^2+h$

where k and h are constants to be determined. First we check that $v( 75 \, mm) = 1.125 \, m/s$ :

$v( 75 \, mm) = k(75 \, mm -75 \, mm) ^2+h = h = 1.125 \, m/s\\\\h= v_{max} = 1.125 \, m/s$

So we found that h was the maximum velocity for the fluid, now we have to determine k, for that we need to make use of the no-slip condition.

$v( 0) = k( -75 \, mm) ^2+ 1.125 \, m/s= 0 \quad (no \, \textendash slip) \\\\k= - \cfrac{ 1.125 \, m/s }{(75 \, mm ) ^2} = - \cfrac{ 1125 \, mm/s }{(75 \, mm ) ^2}\\\\k= - \cfrac{0.2}{mm \times s}$