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Question

Figure shows a $5 k g$ ladder hanging from a string that is connected with a ceiling and is having a spring balance connected in between. A boy of mass $25 k g$ is climbing up the ladder at acceleration $1 m / s^{2}$ . Assuming the spring balance and the string to be massless and the spring to show a constant reading, the reading (in $k g$ ) of the spring balance is: (Take $g = 10 m / s^{2}$ )

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Solution

Draw free body diagram for the man & ladder.

Find the value of tension $T$ .

From the FBD of man,
$N - m g = m a$
$N - (25 \times 10) = (25 \times 1)$
$N = 275$

From the FBD of ladder.
$T - M g - N = 0$
$T - (5 \times 10) - 275 = 0$
$T = 325 N$

Therefore,spring balnce will show $32.5 k g$

Final answer: $32.5$

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