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The Booming Sights of Mount Bromo

With its last eruption in 2007, Mount Bromo is an active volcano in East Java, Indonesia. Although not the highest, it is the most famous peak inside the Tengger Caldera. The picturesque vision of a mountain with belching white sulphurous smoke and a top that has its edges tinged with sulphur is simply breathtaking. The gray volcanic sand inside the caldera often known as the sea of sand only contributes to the scenic imagery of the sight.

The uniqueness of the area is said to be the characteristics reminiscent of a moonscape. The legend behind Mount Bromo is also said to be one of the reasons as to why this eruptive mountain seems to be more famous than others. The legend goes that the mountain gods who were denied the sacrificial 25th child of the 15th century princess named Roro and her husband Joko, sent fire and brimstone to consume the child while demanding an annual ceremony to be performed to pacify the gods. It is said that the performance of the ceremony can be seen up to date.

Apart from the main sight, Mount Bromo it self, Mount Semeru, the tallest mountain in Java and one of the most active volcanoes in Indonesia is a must-see when visiting Tengger Caldera. At the foot of the mountain is a Hindu temple that also attracts many visitors. Mount Batok, situated at the northern centre of the caldera is yet another sight that is worth a visit. Its growing vegetation such as the cemara trees that thrive in volcanic ash is of particular interest to most.

If you’re wondering about accommodation while visiting these volcanic sights, a Surabaya luxury hotel such as the Shangri-La Hotel, Surabaya, provides a comfortable stay with easy access to Mount Bromo and nearby attractions.

About the Author

Pushpitha Wijesinghe is an experienced independent freelance writer. He specializes in providing a wide variety of content and articles related to the travel hospitality industry.

Moment Problem, Help! A crawler crane of 10,000N weight, has a 20-M boom weighing 1000N acting midway....?

of its length, and mounted on a crawler with a track measuring 2 meters between the centers of its track. If the boom is positioned on the side and is to pick up a maximum load of 2000N, what should be the weight and the distance from the midpoint of the crane, of the counterweight such that the crane will not overturn on its side while lifting the load and when there is no load anymore.
If the counterweight is too heavy, the crane will overturn on its side once there is no load to balance it.
It's okay Questor, anyway you have answered.

Sorry I'm late.

From your description, the crane may look like this, with its boom positioned to the side lifting the load.
(hope this serves the purpose)

Wc = the counterweight (?)
X = the distance of the counterweight from the center of the crane
L = Load = 2000N (maximum)
Boom length = 20 M
Wb = Weight of boom = 1000N acting at midpoint(10m from center)
W = weight of the crane = 10,000 N (this could be the body only, excluding the boom and counterweight because these are given separately)

.....................o.
.....Wb..|.../....|....
............|........|..........
......./...|/10M.|.............
...iO=======o-------------[[[[]..Wc
...|............(((((()))))...--.=.=|
.[[]].(Load)|||||||||||||||||||||. =.=|
.............||||||||||||||||||||||||||..=.=|
______[[[]]]___|___[[[[]].__|_______
......Ra./|........|...Rb./|.....|
............|<--1-.>|<-1-->|.....|
............|..........|.........|.==|
...|<---20M.----->|<----X----->|
.......................|
......................|/..W = 10,000 N

Let (Ra) and (Rb) be the reaction of the ground on the tracks of the crane. This reactions are acting on the centers of the track 2 meters apart.

Condition A:
When the boom lifts the maximum Load, then the tracks at Rb will be almost lifted off the ground, so Rb = 0,

Taking summation of moments about Ra (Ra will be the pivot point), so Ra will have no moment.(Clockwise moments--positive, counterclockwise moments--negative)

Considering the system in equilibrium.
M@Ra = 0
0 = Wc(X +1) + W (1) - Wb(10 -1) - L(20-1)
...= Wc(X + 1) + 10,000 (1) -1000(10 - 1) - 2000(20 - 1)
...= Wc (X + 1) + 10,000 - 9000 - 36000

0 = Wc(X + 1) - 35000 <===== equation 1

Condition B
When the load is taken, the crane will have the tendency to overturn towards the side of the counterweight, so, Ra must be not touching the ground, hence, Ra = 0

Considering the system to be in equilibrium
M@Rb = 0
0 = Wc(X - 1) - W(1) - Wb(10 + 1)
...= Wc(X - 1) - 10,000(1) - 1000( 10 + 1)
...=Wc(X-1) - 21,000

Wc = 21,000 / (X - 1) <==equation 2,

Substitute eq. 2 to eq. 1

0 = [21,000 / (X - 1)][X + 1] - 35,000 <==divide by 1000 and simplify
35(X - 1) = 21(X + 1)
35X - 35 = 21X + 21
.........14X = 56

X = 4 M

Substituting this value to eq 2

Wc =21,000/ (4 - 1)
......= 21,000 / 3

Wc = 7,000 N.

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