If I take a weighing scale to the Moon, would it read the same reading as it would on the Earth?
Perhaps. It depends on the design of scale.I’m going to assume our scale is in air at the same pressure on both the earth and the moon, otherwise the answer gets more complex.If you have a scale like this…That uses masses to balance the object, it will give the same answer.If you have a scale like thisthat uses a spring or other mechanism to measure the force exerted by the object you are weighting then no, it won’t read the same.This is because the weight of the object is a force, measured in newtons and it changes depending on how strong gravity is, but the mass, measured in kilograms is a fixed quantity.
Why can a scale for weight go into negatives? Can I get a really simple explanation?
Usually, this is for one of 2 reasons: tare, or dual direction sensor.Tare is a word used to describe zeroing out a scale. The cool thing about this type of scale, is if you needed to measure 100 grams of flour, you could put your empty measuring cup on the scale, press the tare button, and remove your scoop. Now, while the scoop is removed, the scale would read a negative number, which is actually the weight of the missing cup, to let you know you zeroed the scale to the cup. When you put the cup with flour in it on the scale, the only weight being measured is the weight of the flour, and the weight of the cup is ignored.(You could also manually calculate this, but that is more work, so the tare button is really just a time saver)The other option, is that the sensor in the scale could be built to work in both directions. Many scales do not actually measure just weight, but they are really force sensors, being used to measure the force of gravity(weight is actually the force of gravity acting on whatever you are weighing) This means that the scale may also be used for measuring the lifting force of helium balloons or some other things, or could be measuring a pushing force and also a pulling force being placed on it. This would not be true for a balance scale, like those used in a doctors office, or in a laboratory, as these actually compare the mass of a know object to the mass of an unknown object. (These scales measure mass and not force in other words.)
A person has a weight of 600 N standing on a scale in an elevator. The elevator is moving downward with an acceleration of 4 m/s2. What is the apparent weight measured by the scale (in N)?
EDIT: Oops, I read the question wrong. Below is my corrected answer with the elevator accelerating downward.What is the reading on the scale?Simply draw a free body diagram of the man and write Newton’s second law for the free body diagram. A free body diagram shows all the forces acting on the object. In the case of the man, we only have the weight acting down and the contact force from the scale acting up:I’ve also shown the free body diagram for the scale. Note that the reading on the scale ([math]F_{scale}[/math]) is from the force acting down on the scale. Which is also equal and opposite to the force acting up on his feet.For the FBD of the man:[math]\Sigma F_y=ma_y[/math][math]a_y=-4 \frac{m}{s^2}[/math] since it is downward,Since mass = ([math]\frac{weight}{g}[/math]) and assuming positive=up, negative=down, we get:[math]-600N+F_{scale}=(\frac{600N}{9.81})(-4 \frac{m}{s^2})[/math][math]F_{scale}=355 N[/math]
A man, weight = 786 N, stands on a bathroom scale in an elevator. What does the scale read when the elevator i?
(B) When the elevator is moving with a constant speed, its acceleration is zero. If the elevator isn't accelerating, it isn't putting a net force on the man, and it won't affect his weight. The scale will read 786 N as long as the elevator moves at a constant speed. (C) If the elevator is rising and accelerating, its acceleration is +2.11 m/s². The elevator must apply 786 N just to support the man's weight, plus another force to cause his acceleration. If we know the man's mass and his acceleration, we can calculate this net force. F = ma The man's mass can be found from his weight and the acceleration due to gravity, 9.81 m/s²: Fw = mg m = Fw / g = 786 N / 9.81 m/s² = 80.1 kg F = (80.1 kg)(2.11 m/s²) = 169 N So the elevator must supply 169 N in addition to the man's weight to accelerate him upward at 2.11 m/s²; therefore, the scale reads 786 N + 169 N = 955 N (D) If the elevator is accelerating downward at -2.11 m/s², it is no longer fully supporting the man's weight (if it were, he wouldn't be accelerating downward along with it). We already know that the net force on the man must be 169 N, but this time we subtract it from his weight to find what the scale shows: 786 N - 169 N = 617 N I hope that helps. Good luck!
Would weighing an object using two sets of weighing scales give its exact weight?
Weigh means to swing, or dangle. ‘Anchors aweigh’ means that the anchor is lifted up and swung off the side of the ship.Weight = weigh + t in the same way that height = high + t and width = wide + t.Until the industrial era, the only lawful way to determine the weight of a balance was to set the unknown and known weights on pans hanging from a balanced beam, and establish a condition that both pans would dangle free of the bottom.This is actually a torque balance, where we equate gml = GML, where capitals refer to the known pan, g is gravity, m is mass, and l is the length from the pivit.g = G to one part in 100,000,000 or better. In any case, it’s more certain than other effects can disrupt this.m = m’ - kTv where v is the volume of the mass, t is the temperature of the air, and k is some constant. Again, k = 1/2000000 for water. It has a minimal effect, but much larger than G vs g.L is only known imperfectly, and while temperature can increase the arms, it cancels out. On the other hand L/l is bettter determined by reversing the weights, and it is for this reason that equal-arm balances are held to be the most lawful weights.
Why does a weighing machine read more than my weight if I jump from some height?
A2A. A weighing machine is really just some sort of force meter, typically based on either a spring or a balance beam. It doesn't directly measure the force of gravity on you, it measures the force you apply to it (which per Newton's Third Law is the opposite of the force it applies to you).Now if you use it according to the manual (which hopefully will say common-sense things like, "Stand still, don't be wearing or carrying anything else, and don't lean or otherwise push on anything else."), then first of all you will not be accelerating. (That's implicit in the "stand still" part.) Therefore, per Newton's Second Law, the net force on you will be zero.And since you're not carrying or pushing on anything else, the only forces on you are gravity downward and the force upward from the weighing machine. Since these have to be equal but opposite to sum to zero, the weighing machine is effectively registering your weight.However if you jump onto the weight machine, it's telling you the sum of you weight plus the force required to decelerate you from falling to stationary. This latter contribution is proportional to your mass but also to how fast you stop, which could vary over a wide range.
What does the scale read during the acceleration? Any help would be very appreciated!?
OK so first let's calculate the woman's weight at rest. F=mg 74 * 9.8 = 725.2 next calculate the weight of the entire elevator 815 + 74 = 889 889 * 9.8 = 8712.2 the net force of 9830 up and 8712.2 down gives 1117.8 total up force. next calculate the acceleration of the elevator upwards. 1117.8 = 889a a = 1117.8/889 now calculate the additional force felt by the woman due to this acceleration. F = 74 * (1117.8/889) F = 93 now calculate total force applied to the scale. Initial weight plus additional weight applied. 725.2 + 93 = 818.2 N you can then convert this weight into the weight of any unit you prefer. now on the other hand if this scale is measuring mass (there is no mention if it's a weight or mass scale) then the scale will remain the same since the woman's mass does not change.