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Unformatted text preview: or Pcos18-T-Wcos53=ma #3 girl Scale Assuming the girl accelerates downward… and using the standard choice of coordinate system, this gives: T-W g = m g (-a) and T+N-W bl = m bl (a) If the scale is to read zero, the normal force must be zero. So we have T-W g = m g (-a) and T-W bl = m bl (a) If the block is to remain in contact with the scale, a for the block is zero. So we have T-W g = m g (-a) and T-W bl = 0 or T=W bl Combining these two gives: W bl-W g = m g (-a) or M bl (g) – m g (g) = m g (-a) or g ( M bl- m g )/m g =-a So g (75-50)/(50) = -a -a = 0.5 g so a = -0.5 g The minus sign means that I chose the wrong direction for the acceleration in my initial assumption, so N S on Bl W Bl N T on S W S F Bl on S (N Bl on S ) T W g N S on B W Bl T a is 0.5 g UPWARD....
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- Spring '08
- Force, Normal Force, MBL, 2. 3. 4. 5. 6. 7. 8, C E C B