A meter stick (of length 1.00 m) weighs 1.40 N (uniform distribution) and is hel

Question

A meter stick (of length 1.00 m) weighs 1.40 N (uniform distribution) and is held in static equilibrium by two vertically directed forces, one located at x =0.00 cm and another located at x =5.00 cm. No other forces (besides gravity) act on the meter stick. Calculate both forces (no signs for the magnitudes) The force at x = 5.00 cm is now moved to x = 3.00 cm. In order to maintain static equilibrium, both forces must be increased in magnitude, only one of the forces must be changed. one of the forces must be increased in magnitude and the other decreased in magnitude, both forces must be decreased in magnitude, neither force must be changed.

Explanation / Answer

Here ,

for finding the force at x = 5 cm = 0.05 m

balacing the moment about x = 0

F * 0.05 - 1.4 * 0.50 = 0

F = 14 N

the force at x = 5 cm is 14 N upwards ,

Now, for balancing y -forces

14 + F = 1.4

F = - 12.6 N

force at x = 0 cm is 12.6 N downwards.

Here ,

if the x = 5 cm is moved to 3 cm ,

>>both the forces must be increased in magnitude

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