A uniform 20.0-kg, 6.00-m long beam is supported in equilibrium by two cords fro

Question

A uniform 20.0-kg, 6.00-m long beam is supported in equilibrium by two cords from the ceiling, as shown in Fig.i. Determine the forces F_A and F_s applied by the cords to the beam. A small planet having a radius of 1000 km exerts a gravitational force of 100 N on an object that is 500 km above its surface. If this object is moved 500 km farther from the planet, what will the gravitational force on it be? A 8.1-kg solid object, made of metal whose density is 2700 kg/m^3, is suspended by a cord. When the object is immersed in water (of density 1000 kg/m^3) and is held at rest, what Is the tension in the cord? An incompressible fluid flows steadily through a horizontal pipe that has a change in diameter. The fluid speed at a location where the pipe diameter 8.0 cm is 1.3 m/s. What is the fluid speed at a location where the diameter has narrowed to 4.0 cm?

Explanation / Answer

7.Refer the figure below,

Applying Newton’s second law vertically,

F(A)+F(B) = mg ---------------(1)

Applying law of conservation of torque at F(B),

F(B)*0 - mg*3 + F(A)*4.5 =0

- mg*3 + F(A)*4.5 =0

-20*9.8 + F(A)*4.5 =0     => F(A) = 43.56 N

From (1)

F(B) = mg – F(A) = 20*9.8 – 43.56 = 152.44 N

8.Use equation,

F= GMm/r^2

It indicates F 1/r^2

Thus we can write,

F2/F1 = r1^2/r2^2

F2 = F1*(r1^2/r2^2) = 100*[(10^6 +5*10^5)^2 /(10^6 +10^6)^2] = 56.25 N

9. volume of the object V = m/ = 8.1/2700 = 0.003 m^3

Now applying Newton’s second law to the immerged object

T +Fb – mg = 0      ( Fb = buoyant force)

T = mg – Fb = mg - Vobject *water*g = 8.1*9.8 – 0.003*1000*9.8 = 49.98 N

10. By continuity equation

A1v1 = A2v2 =

r1^2*v1 = r2^2*v2

r1^2*v1 = r2^2*v2

v2= v1*(r1^2/r2^2)= 1.3*(0.04^2/0.02^2) = 5.2 m/s

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