Three boards, each 50 mm thick, are nailed together to form a beam that is subjected to a 1200-N vertical shear. Knowing that the allowable shearing force in each nail is 700 N, determine the largest permissible spacing s between the nails.

Using equation

qs=F

q=VQ/I   V=1200

q=998.45 N

qs=F

s=F/q

s=700/998.45

spacing=s=0.701 m

The question is incomplete. The mass of the board is needed in order to calculate the moment of inertia of the beam. 91.67 * 10⁶mm⁴ is used as the moment of inertia in the calculation below. You can always substitute to get your answer.

The spacing is  95.1mm

Explanation:

Given'

vertical shear V = 1200N

Shearing force = 700N

Thickness of each board = 50mm

The spacing can be calculated using the formula;

F = q * s1

but, q = VQ/I2

Therefore,  F = VQ/I  * s 3

where,

s = spacing

v = vertical shear

f = shearing force

I = moment of inertia

Making s subject formula in equation 3, we have

s = FI/VQ4

but Q = YA

A = area = 50 *(150)

Y = 75

therefore, Q = 75 *50*150

Q= 5.625* 10⁶mm⁴

Substituting into equation 4, we have

s = 700 * 91.67 *10⁶) / (1200*5.625* 10⁶)

   =  6.42*10¹⁰ /6.75*10⁸

  = 95.1mm

potential energy is directly proportional to height.

more height = more potential energy.

greatest increase in height = greatest increase in potential energy

where are the points with the greatest increase in height ?

has to be c-e or d-e .

the answer is b, have negligible mass