Alan

This can be done as follows:

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sin is 1/ csc so if csc theta = 8 then sin theta = 1/8

1 - What is Rc ? is this the shear force ?? 

In the free-body diagram for CD, R_C is the net force on CD due to AC (i.e. AC is helping to support CD and keep it from going down).

In the free-body diagram for AC, R_C is the reaction of CD on AC (Newton's 3rd law of motion).

2- Is RE is the reaction of the support in beam BE ? 

Yes, in free-body diagram BE, R_E is the force of the support at E on BE.  

3- What is RB in the beam BE ? 

In free-body diagram BE, R_B is the force of AC on BE (i.e. AC is trying to pull BE down).

In free-body diagram AC, R_B is the reaction force of BE on AC (i.e. BE is trying to keep AC up - Newton's 3rd law again).

Hope this helps.

The geometric mean of n numbers is given by the n'th root of the product of all the numbers, which is how Guest got his/her result.

The gravitational attraction is given by \(F=G\frac{m_1m_2}{r^2}\)

G = 6.6742*10^-11*m^3/(kg*sec^2)

The m's are masses and r is the distance between them.

F = 6.6742*10^-11 x (3.1x3.1)/(0.26^2) = 9.49x10^-9 N

The units are Newtons - a force (not m/sec^2 which is acceleration)

Likewise!

Guest is correct.  Note that the speed of light he/she used has units of m/s  (and Hertz = s^-1).

tan(pi/2) is infinite, so log(tan(pi/2)) is also infinite.

Try it as follows:

Numerator 

sqrt(64,5) → sqrt(2*32,5) → sqrt(2,5)*sqrt(32,5) → sqrt(2,5)*sqrt(2^5,5) → sqrt(2,5)*2

sqrt(2,5) is 2^(1/5), so sqrt(64,5) → 2*2^(1/5)

Hence sqrt(64,5)*2^(-1/5) → 2*2^(1/5)*2^(-1/5) → 2

Denominator

sqrt(8) → (sqrt(2)*sqrt(4)) → 2*sqrt(2)

Numerator/Denominator → 1/sqrt(2)

Raise this to the power -4:   (1/sqrt(2))^-4 → sqrt(2)^4 → (sqrt(2)^2)^2 → 2^2 → 4

I think the following is probably what is needed (but you should check it carefully!).

Note:  The force on top of BE should be labelled R_E not R_D (which was a typo by me).  It will be the same in magnitude as R_B. 

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