(II) Suppose the conveyor belt of Example 9–20 is retarded by ...

Question

(II) Suppose the conveyor belt of Example 9–20 is retarded by a friction force of 150 N. Determine the required output power (hp) of the motor as a function of time from the moment gravel first starts falling (t = 0) until 3.0 s after the gravel begins to be dumped off the end of the 22-m-long conveyor belt.

Accepted Answer

Hello, fellow physicists today, we're to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. Consider a situation in which a manufacturing assembly line is in operation in this setup. A conveyor belt maintains a consistent speed of 2.5 m per second while grains are deposited onto it from a hopper at a rate of 65 kg per second. If the conveyor belt experiences a friction force of 180 newtons causing it to slow down, calculate the requisite output power in units of horse power that the motor experiences over time. This analysis should commence from the instant the grains are initially deposited at T equals zero seconds and extend until 4.0 seconds after the grains commence sliding off the end of the 25 m long conveyor belt. Awesome. So it appears that our final answer that we're ultimately trying to solve for is we're trying to figure out what their required power output power. So we're trying to figure out what their required output power is in units of horsepower that this motor experiences over time, considering all of the conditions provided to us by the problem itself. So we're ultimately trying to figure out what they're required, which is the requisite power output power to be exact the output power and units of horsepower. So that's what we're ultimately trying to solve for is this requisite output power value? Awesome. So before we read off our multiple choice answers, let's quickly look at the diagram that's provided to us by the prom itself. So we have like this thing that shaped like half of an hour glass, an old school hourglass. And it's our hopper. Our hopper is this one that's like a rectangle or like a rectangular shape with a little opening where it's depositing grains where the grains are, these black circles or black filled in circles and they're dropping onto this conveyor belt which is moving at some speed which we know that they're dropping onto the belt at a speed or a rate of I should say specifically a rate of 65 kg per second. So the grains are being spit out, they're being dropped from the hopper onto the conveyor belt which is moving at some velocity T vector vector V velocity to the left. So this conveyor belt and then you see a little circle here at the bottom of our conveyor belt. So that's the little rolling cylinder that this belt is being rolled over. So it's being rolled around. So the belt is moving, rotating to the left. Awesome. So now that we have a visualization of what's going on in this particular problem, let's read off our multiple choice sensors to see what our final answer might be. And let us note that they're all in the same units of horsepower HP. So A is 1.2 B is 0.87 C is 0.82 and D is 0.46. OK. So with that in mind, we 1st, 1st. So first off, we need to note that the external force on the belt is the force supplied by the motor and the oppositely directed force of the friction. So now we must treat our scenario provided to us by the promise of to be a system containing tiny masses of D capital M to represent the grains that travel with a velocity of U which is about to enter the mass capital M that represents the conveyor belt. Let us note that an infinitesimally small time DT later capital. So some infinitely time, DT later D capital M, we'll combine with mass capital M. Let us refer to this situation as the collision. Therefore, our large mass will change its mass from capital M to capital M plus D capital M. So note that there will be a change in momentum. Therefore, we can recall and write and let's call this equation one that DP is equal to capital M plus D capital M. And we need to multiply this by V plus DV. And we need to subtract this from capital M V plus U multiplied by DM. OK. So let us note that the mass of the total system, which is the conveyor belt plus the grains is, and we could write this as MT is equal to capital M plus DM or specifically lowercase M subscript T is equal to capital M plus D capital M. So the velocity of the whole system after the collision can be written as VW is equal to V plus DV. And let us also note that MV is equal to the initial momentum of the conveyor belt. And let us also note that U multiplied by D capital M is equal to the initial momentum of the grains. So considering equation one, we can write that DP is equal to capital M multiplied by DV plus V multiplied by D capital M plus D capital M multiplied by DV minus U multiplied by D capital M. So let us note at this point that the term D capital M multiplied by DV is the product of two differentials and will become zero once we divide it by D lowercase T. Thus, we can write that DP is equal to capital M multiplied by DV plus V multiplied by D capital M minus U multiplied by D capital M. So now we will have to take the sum of all the external forces acting on our particular system, which we can write as capital F external is equal to DP by DT, which we can go ahead and write that F external is equal to capital M multiplied by DV plus V multiplied by D capital M minus U multiplied by D capital M divided by, are actually not divided by anything. Thus, we can go ahead and write that F external is equal to capital M multiplied by DV, divided by DT minus U minus V multiplied by D capital M divided by DT. So we can simplify again and write that F external is equal to capital M. And we need to take capital M and we need to multiply it by DV divided by DT and we need to subtract it from the relative. So lowercase R is gonna denote relative V relative multiplied by D capital M divided by DT. Thus, we can go ahead and write that M or specifically capital M multiplied by DV, divided by DT is equal to F external plus the relative multiplied by capital or DD capital M by DT. And let us note that DV by DT is equal to the acceleration of the belt which is equal to zero. So let's make a little note here off to the side that DV by DT is equal to the acceleration of the belt which is 0 m per second squared. So since the belt moves at a constant speed. So this is because the belt moves at a constant speed. And let us note that the relative is the relative velocity and VR our relative velocity is equal to U minus V, which is equal to zero minus V which is equal to negative V and D capital M by DT is the rate at which the grains are deposited. So DM by DT which once again, this is the rate at which the grains are deposited is equal to 65 kg per second. Awesome. And that F external is the external force on the system. And fext is equal to the force on the motor, which we're gonna call F subscript M plus the force due to friction, which we're gonna denote friction as lowercase F. And let us note that our friction force F subscript lowercase F is equal to negative 180 newtons. So note that the negative sign indicates the direction of friction and since the direction of friction is in the opposite direction of the motion, so that's why it's negative. Thus, we can go ahead and write that capital M of zero is equal to F um plus FF minus V multiplied by D capital M by DT. Thus, we can go ahead and write that FM is equal to V multiplied by D capital M by DT minus SS which when we plug in our known variables, we can go ahead and write that this is all equal to 2.5 meters per second, multiplied by 65 kg per second minus negative 180 newtons. Which when we perform that calculation, we will get 342 0.5 newtons when we round to one decimal place. Thus, the power output needed from the motor is determined by multiplying the force by the velocity, which we can write this equation as P motor, which is the power of the motor is equal to FM, multiplied by V which is equal to our FM value, which we just determined to be 342.5 newtons. And we need to multiply this by two 0.5 meters per second. So let me plug that indoor calculator. We should get 856.25. Let me round to two decimal places watts. But now we need to convert watts to horsepower. So we need to take 856.25 watts. And using dimensional analysis, we need to note that in one horse power there is 746 watts. So when we plug that into our calculator, we will get when we round to one decimal place, 1.2 horsepower. And that is our final answer. Hooray, we did it. So we need to note that as the grains fall off of the conveyor belt, it does not experience horizontal acceleration from the belt. Therefore, there will be no additional force interaction with the belt. However, each new batch of falling grains still needs to be accelerated and will maintain a requirement of needing a constant power source. Thus, the power will remain constant from T equals zero to T equals 4.0 seconds. And that's it we've solved for this problem. So looking at our multiple choice answers, the correct answer has to be the letter A 1.2 horsepower. Thank you so much for watching. Hopefully that helped and I can't wait to see you in the next video. Bye.

Key Concepts

Power

Friction Force

Kinematics of the Conveyor Belt

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