Answer : 413.44N
Here it is given that an elevator is moving down with an acceleration of 3.36 m/s² . And we are interested in finding out the apparent weight of a 64.2 kg man . For the diagram refer to the attachment .
- From the elevator's frame ( non inertial frame of reference) , we would have to think of a pseudo force.
- The direction of this force is opposite to the direction of acceleration the frame and its magnitude is equal to the product of mass of the concerned body with the acceleration of the frame .
- When a elevator accelerates down , the weight recorded is less than the actual weight .
From the Free body diagram ,
[tex]\sf\longrightarrow Weight = mg - ma \\[/tex]
[tex]\sf\longrightarrow Weight = m ( g - a ) \\[/tex]
- Mass of the man = 64.2 kg
[tex]\sf\longrightarrow Weight = 64.2( 9.8 - 3.36) N\\[/tex]
[tex]\sf\longrightarrow Weight = 64.2 * 6.44 N\\[/tex]
[tex]\sf\longrightarrow \underline{\boxed{\bf Weight_{apparent}= 413.44 N }} \\[/tex]
