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What does the scale read if the cab is stationary or moving upward at a constant 0.50 m/s? |
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Answer» Solution :For any constant velocity (zero or otherwise), the acceleration a of the passenger is zero. Calculation: Substituting this and other known VALUES into EQ. 5-38, we find `F_(N)=(72.2" kg")(9.8" m"//"s"^(2)+0)=708" N"`. This is the weight of the passenger and is equal to the magnitude `F_(g)` of the gravitational force on him. ( c ) What does the scale read if the cab accelerates upward at `3.20" m"//"s"^(2)` and downward at `3.20" m"//"s"^(2)` ? CALCULATIONS: For `a=3.20" m"//"s"^(2)," Eq. "5-38` gives `F_(N)=(72.2" kg")(9.8" m"//"s"^(2)+3.20" m"//"s"^(2))` `=477" N"`. For an upward acceleration (either the cab.s upward speed is increasing or its downward speed is decreasing), the scale reading is greater than the passenger.s weight. That reading is a measurement of an apparent weight, because it is made in a noninertial frame. For a downward acceleration (either decreasing upward speed or increasing downward speed), the scale reading is less than the passenger.s weight. (d) During the upward acceleration in in part (c), what is the magnitude `F_("NET")` of the net force on the passenger, and what is the magnitude `a_("p, cab")` of his acceleration as measured in the frame of the cab? Does `vec(F)_("net")=m vec(a)_("p, cab")` ? Calculation: The magnitude `F_(g)` of the gravitational force on the passenger does not depend on the motion of the passenger or the cab, so, from part (b), `F_(g)` is 708 N. From part (c), the magnitude `F_(N)` of the normal force on the passenger during the upward acceleration is the 939 N reading on the scale. Thus, the net force on the passenger is `F_("net")=F_(N)-F_(g)=939" N"-708" N"=231" N"` during the upward acceleration. However, hsi acceleration `a_("p. cab")` relative to the frame of the cab is zero. Thus, in the noninertial frame of the accelerating cab, `F_("net")` is not equal to `ma_("p, cab")`, and Newton.s second law does not hold. |
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