Question:
The hoisting system of a drilling rig contains seven ideal sheaves with hook load of \( 3.0 \times 10^5 \) kg.
The static derrick load is .......... × \( 10^5 \) kg (rounded off to one decimal place).
The hoisting system of a drilling rig contains seven ideal sheaves with hook load of \( 3.0 \times 10^5 \) kg.
The static derrick load is .......... × \( 10^5 \) kg (rounded off to one decimal place).
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To calculate the derrick load, consider the number of sheaves and use the formula that accounts for the distribution of the load across the sheaves.
Updated On: Jan 30, 2026
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Correct Answer: 3.9
Solution and Explanation
The formula for the static derrick load in such systems is given by:
\[ {Derrick Load} = \frac{{Hook Load} \times {Number of Sheaves}}{2} \] Given:
- Hook load per sheave: \( 3.0 \times 10^5 \, {kg} \)
- Number of sheaves: 7
Substituting the values into the formula:
\[ {Derrick Load} = \frac{(3.0 \times 10^5) \times 7}{2} = \frac{2.1 \times 10^6}{2} = 1.05 \times 10^6 \, {kg} \] Thus, the static derrick load is approximately \( 1.05 \times 10^6 \, {kg} \), or \( 1.05 \times 10^5 \, {kg} \) when rounded off to one decimal place.
The static derrick load is between \( 3.9 \times 10^5 \, {kg} \) and \( 4.1 \times 10^5 \, {kg} \).
\[ {Derrick Load} = \frac{{Hook Load} \times {Number of Sheaves}}{2} \] Given:
- Hook load per sheave: \( 3.0 \times 10^5 \, {kg} \)
- Number of sheaves: 7
Substituting the values into the formula:
\[ {Derrick Load} = \frac{(3.0 \times 10^5) \times 7}{2} = \frac{2.1 \times 10^6}{2} = 1.05 \times 10^6 \, {kg} \] Thus, the static derrick load is approximately \( 1.05 \times 10^6 \, {kg} \), or \( 1.05 \times 10^5 \, {kg} \) when rounded off to one decimal place.
The static derrick load is between \( 3.9 \times 10^5 \, {kg} \) and \( 4.1 \times 10^5 \, {kg} \).
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