What weight does the scale report when 440 kg of sand have already been added?

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Homework Help Overview

The problem involves a pickup truck with a mass of 2400 kg that is weighed on a scale while sand is added at a rate of 150 kg/s from a height of 3.0 m. At the moment when 440 kg of sand has been added, the question is what weight the scale reports.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the relevance of impulse and kinematics to the problem. Questions arise about the velocity of the sand after falling and the rate of change of momentum when it impacts the truck bed. There are attempts to apply kinematic equations to find the velocity of the sand as it hits the bed.

Discussion Status

Participants are exploring various interpretations of the problem, particularly focusing on the calculations related to the velocity of the sand and its momentum. Some guidance has been provided regarding the calculations of momentum and the forces involved, but no consensus has been reached on the final weight reported by the scale.

Contextual Notes

There is an ongoing discussion about the assumptions regarding the initial velocity of the sand and the effects of gravity on its motion. The problem constraints include the mass of the truck and the rate at which sand is added, which are central to the calculations being discussed.

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Homework Statement



A pickup truck has a mass of 2400 kg when empty. It is driven onto a scale, and sand is poured in at the rate of 150 kg/s from a height of 3.0 m above the truck bed. At the instant when 440 kg of sand have already been added, what weight does the scale report?

Homework Equations



$$ J=\Delta p=mv_1-mv_0 $$
$$ w=mg $$

The Attempt at a Solution



I can't quite see what this has to do with impulse, and whether or not kinematics have to be used in this at all or not.
 
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rountree85 said:

Homework Statement



A pickup truck has a mass of 2400 kg when empty. It is driven onto a scale, and sand is poured in at the rate of 150 kg/s from a height of 3.0 m above the truck bed. At the instant when 440 kg of sand have already been added, what weight does the scale report?

Homework Equations



$$ J=\Delta p=mv_1-mv_0 $$
$$ w=mg $$

The Attempt at a Solution



I can't quite see what this has to do with impulse, and whether or not kinematics have to be used in this at all or not.

What is the velocity of the sand after it has dropped the 3 m? What is the rate of change of momentum of the stream of sand when it hits the truck bed?
 
Well, would we use the fact that initial velocity is zero and final is the unknown, the acceleration downward is that of gravity, which is -9.8 m/s^2, and the displacement is 3.0 m so you would use $$ v^2=v_0^2+2a\Delta y $$ to find the velocity?

And, if the initial velocity is zero, then the change of momentum would just be the mass times the velocity calculated above, correct?
 
rountree85 said:
Well, would we use the fact that initial velocity is zero and final is the unknown, the acceleration downward is that of gravity, which is -9.8 m/s^2, and the displacement is 3.0 m so you would use $$ v^2=v_0^2+2a\Delta y $$ to find the velocity?

And, if the initial velocity is zero, then the change of momentum would just be the mass times the velocity calculated above, correct?

Correct on getting the velocity of the sand when it hits the bed. The downward velocity of the sand then goes to zero. If the rate of sand flow is 150 kg/sec, and its velocity goes from v to zero when it hits the bed, what is the rate of change of momentum of the sand which hits the bed?
 
So solving for v by putting in

$$ v^2=2(9.8 m/s^2)(3.0 m) $$

we get 7.7 meters per second.

So, if I'm correct, the formula for rate of change of momentum would be the force, which is equal to the change in momentum, Δp, over the change in time, which is Δt. We can calculate change in time by taking the mass of sand, which is 440 kg, and dividing it by the rate, 150 kg/s, to get 2.9 s. So the change in momentum at the top is just mass of the pickup truck, 2400 kg, plus the 440 kg, times the velocity I calculated, 7.7 m/s, and since it goes to zero, the final momentum is zero.
 
rountree85 said:
Well, would we use the fact that initial velocity is zero and final is the unknown, the acceleration downward is that of gravity, which is -9.8 m/s^2, and the displacement is 3.0 m so you would use $$ v^2=v_0^2+2a\Delta y $$ to find the velocity?

And, if the initial velocity is zero, then the change of momentum would just be the mass times the velocity calculated above, correct?

rountree85 said:
So solving for v by putting in

$$ v^2=2(9.8 m/s^2)(3.0 m) $$

we get 7.7 meters per second.

So, if I'm correct, the formula for rate of change of momentum would be the force, which is equal to the change in momentum, Δp, over the change in time, which is Δt. We can calculate change in time by taking the mass of sand, which is 440 kg, and dividing it by the rate, 150 kg/s, to get 2.9 s. So the change in momentum at the top is just mass of the pickup truck, 2400 kg, plus the 440 kg, times the velocity I calculated, 7.7 m/s, and since it goes to zero, the final momentum is zero.

Not quite. If the rate at which the sand is hitting the bed is 150 kg/sec, and its velocity decreases from 7.7 m/s to zero when it hits the bed, the rate of change of momentum for the sand stream hitting the bed is 150 x 7.7 = 1155N. The mass of the truck is 2400 kg, and, when 440 kg of sand is already at rest in the bed, the total mass of truck + sand at rest = 2840 kg. Therefore, the total force measured by the scale = 2840 (9.8) + 1155 Newtons.