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

In summary, a pickup truck with a mass of 2400 kg when empty is driven onto a scale. Sand is poured in at a rate of 150 kg/s from a height of 3.0 m above the truck bed. After 440 kg of sand is added, the weight reported by the scale is equal to the sum of the weight of the truck and the force exerted by the sand stream hitting the bed. This is calculated by adding the weight of the truck and the weight of the sand at rest, and the force of the sand stream hitting the bed, which is equal to the product of the mass of sand added per second and the velocity of the sand when it hits the bed.
  • #1
rountree85
3
0

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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  • #2
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?
 
  • #3
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?
 
  • #4
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?
 
  • #5
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.
 
  • #6
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.
 

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

What is an impulse and weight problem?

An impulse and weight problem is a type of physics problem that involves calculating the force or change in momentum caused by an object's weight and its interaction with another object or surface.

How is impulse and weight related to Newton's laws of motion?

Impulse and weight are closely related to Newton's laws of motion, specifically the second and third laws. The second law states that force is equal to mass times acceleration, and the third law states that for every action, there is an equal and opposite reaction. In an impulse and weight problem, these laws can be used to calculate the force and change in momentum of an object.

What are some real-life examples of impulse and weight problems?

Impulse and weight problems can be found in many real-life situations, such as a person jumping off a diving board, a car crashing into a wall, or a ball bouncing off a surface. In each of these scenarios, the weight and motion of the object interact with another object or surface, causing a change in momentum.

How can impulse and weight problems be solved?

To solve an impulse and weight problem, one must first identify the two objects or surfaces involved and determine their masses and velocities. Then, Newton's second law can be used to calculate the force, and the third law can be used to determine the change in momentum. Other principles, such as conservation of energy, may also be applied depending on the specific problem.

Why are impulse and weight problems important in science and engineering?

Impulse and weight problems are important in science and engineering because they help us understand the behavior of objects in motion and how forces act upon them. This knowledge is crucial in fields such as mechanics, aerospace engineering, and biomechanics, where understanding the effects of weight and impulse is essential for designing and building structures and machines.

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