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4-1.Newton's Laws of Motion
hard
The pulleys in the diagram are all smooth and light. The acceleration of $A$ is a upwards and the acceleration of $C$ is $f$ downwards. The acceleration of $B$ is
A
$\frac{1} {2} (f-a) $up
B
$\frac{1} {2}(a + f)$ down
C
$\frac{1} {2}(a + f)$ up
D
$\frac{1} {2}(a - f)$ up
Solution
$l_{1}+2 l_{2}+l_{3}=$ constant ltbritgt $\frac{d l_{1}}{d t}+2 \frac{d l_{2}}{d t}+\frac{d l_{3}}{d t}=0$
$\frac{d^{2} l_{1}}{d t^{2}}+2 \frac{d^{2} l_{2}}{d t^{2}}+\frac{d^{2} l_{3}}{d t^{2}}=0$
$\Rightarrow-a+2 a_{B}+f=0$
$a_{B}=-\frac{1}{2}(f-a) \Rightarrow a_{B}=\frac{1}{2}(f-a)up$
Standard 11
Physics
