MathRevolution wrote:
There are three pulleys connected to one another. The one on the left has a radius of 8, has number 1,2,3,4 marked clock-wise (arrow is pointing at 0). The one on the right has radius of 12 and has 1,2,3,4 marked clock-wise (again, arrow pointing to 0). The one at the bottom is a pulley with a radius of 4. If the bottom pulley turns 1&1/2 times anti-clockwise, what is the sum of the numbers that the arrows of the two pulleys on the upper part of the figure would be pointing at after 1&1/2 turns?
A. 3
B. 4
C. 5
D. 6
E. 7
*An answer will be posted in 2 days.
since there is NO 0, i believe you mean 1, when you right arrow pointing at 1...
The Q depends on circumference...
the lower pulley has a radius 4 and will travel \(1.5 ( 2pi*4)=12*pi\) in 1\(\frac{1}{2}\)rotations...
lets see the effect on the other two pulleys as the all will travel same distance during this interval...
1)The circumference of pulley of radius 8 is \(2*pi*8 = 16*pi..\)
so the pulley will move \(\frac{12*pi}{16*pi} = \frac{3}{4}\)of the rotation, so the arrow will cover 1 to 2, 2 to 3 and 3 to 4... AND will point at 4..
1)The circumference of pulley of radius 12 is \(2*pi*12 = 24*pi..\)
so the pulley will move \(\frac{12*pi}{24*pi}= \frac{1}{2}\) of the rotation, so the arrow will cover 1 to 2 and 2 to 3... AND will point at 3..
\(SUM = 4+3 = 7\)
E
Hi, thanks for the explanation. Can you also clarify why the upper two pulleys move clockwise if the bottom pulley spins counter-clockwise? I presumed that all pulleys would move counter clockwise and hence got 5 ==> Left pulley goes from 1-4-3-2 and right pulley goes 1-4-3, the sum of both being five.