Two people carry a heavy electric motor by placing it on a light ...

Question

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 N. (a) What is the weight of the motor, and where along the board is its center of gravity located?

Accepted Answer

Hey, everyone in this problem, a crate is placed on a massless plank of three m in length transported by two workers. One worker applies a 300 Newton vertical force at the left hand end of the plank, the other applies a 500 Newton force at the right hand end of the plank. Okay. And we're asked to find the weight of the crate and the position of the center of gravity. Okay. Alright. So let's draw a little diagram first. So we have our plank that's massless. Okay. On the left hand end, we have the force from the first worker. We'll call it F one. On the right hand end, we have this upward vertical force from the second worker F two. And we have our crate placed somewhere here on our plank and it is going to exert a downward force of weight and we'll call it W C the weight of the crate. We know that this is a three m long plank. So the distance between F one and F two here is three m. Alright. And then the plank is massless case. We don't have to worry about the weight of the plank acting downwards. Alright. So the first thing we're trying to do is we're trying to figure out the weight of the crate. Know what we have here is an equilibrium situation, okay. We have no net external forces acting on our system and we have no net external torques acting on our system. Okay. So what this tells us is that the sum of the forces is going to be equal to zero and we're gonna take our coordinate system here. And if the sum of the forces is zero, this gives us two equations. Okay, we have the sum of the forces in the X direction is going to be zero and the sum of the forces in the Y direction is also going to be zero. Alright, so let's start with this. Okay. Now, I also mentioned that we have no net external torques that's going to give us another equilibrium condition that some of the torques is equal to zero. Okay. But we're gonna start with some of the forces. Now, some of the forces in the extraction, we don't actually have any ex forces acting and we're trying to find the weight of the crate which is in the Y direction. So we're gonna start with some of the forces in the Y direction equal to zero. Well, what forces do we have in the positive Y direction? We have the force F one and the force F two. Okay. Those forces those vertical forces that the workers apply at either end of the plank. Okay. And then we have in the negative Y direction, the weight of the crate, All of these are going to be equal to zero. And what you'll see is that we know F1, we know F two. So we can go ahead and find the weight of the crate from this equation. Alright. So the weight of the crate is just going to be equal to the sum of those two forces. Now we know that these forces are 300 newtons. Okay. In 500 newtons, one worker applies a force of 300 newtons. One worker applies a force of 500 newtons which tells us that the weight of the crate is going to be 800 newtons. Alright. So we have the weight of the crate and if you were working on a multiple choice question on a test, you could look at this these answer choices and see that we have either answer C or D okay because we have found 800 newtons. So let's do part two um and figure out the position of the center of gravity. Now we can do this using our torque equation. Okay. For a torque are some of torques. Okay. We need to choose a pip, let's go ahead and choose the left end as a pivot if we choose this as a pivot. Okay. When we're talking about distances or the radius are from the pivot. Okay. Well, then the distance to the crate center of mass, okay. We're gonna call X. Alright. So let's get to our torque equation and see what we can find. Alright. So part two, we're gonna use our second equilibrium condition the some of the torque zero. Okay. Now, if we're using, let's go back to our diagram. If we're using The point on the left here as our pivot, then the force acting right here. F1 is not going to contribute to our torque. Okay. So we only need to worry about the force from the weight of the crate and then this f to force. So thinking about this F to force okay. If you imagine pushing up kind of on this plank, this is going to cause counterclockwise rotation. So the torque from F two is going to be positive and for the weight of the crate, okay. This is a downward force. If you imagine pushing down on this plank, this is going to cause clockwise rotation of the plank. Okay. So this is going to be a negative torque. Alright. So we're gonna have the torque and we'll call it T to the torque related to F to minus the torque related to the crate is going to be equal to zero. What is torque? Torque is R F sine of the angle? Okay. Where R is the distance from the pivot to the center of mass of the object or to the force that we're looking at okay. F is the magnitude of the force and then data is the angle between the axis, a rotation and our force. Okay. So for T two, we get that the distance okay is gonna be the distance of that whole plank because our pivot is on the far left end in this forks acts on the far right end. Okay. So that's gonna be three m, the magnitude of the force which we know is F two times sine of the angle. And these are perpendicular to the force F two X perpendicular to our plank. And so this is gonna be sign of 90 degrees. Now for the crate, well, the distance is going to be X okay. We call the X in our diagram. So we'll put it in blue here just like we did in our diagram, that distance is going to be X okay. The magnitude of the forces W C And again, this acts perpendicular early to the plank. So we get sine of 90°. Okay. Alright. So we divided by sine of 90° and we're looking for X. Okay. So we see that X W C is going to be equal to three m times F two looking for X and we actually know all of these values. Okay. We just found W C. So we have X times 800 newtons is equal to three m times F two and the force F two is gonna be the force applied on the right side and the worker on the right hand side applies a 500 Newton force. Okay. So we have 500 newtons. So take X we have three m, 500 newtons, we get 1500 newton times meter divided by 800 newtons in the unit of Newton cancels And we get an x value of 1. m. Alright. So we go back up to our diagram. What this tells us is the distance from this left end. A two The centre of gravity of the crate is 1.875 m. Okay. All right. So looking at our possible answer choices, okay, we found that the weight of the crate is 800 newtons and the position of the center of gravity is 1.875 m from the left. And so we have answer choice D and just be careful. Okay. In this case, they've said that the distance is measured from the left. Okay. So we chose the pivot from the left, just be careful if this is from the right. You either want to choose your pivot from the right or you're going to need to kind of subtract from the total distance in order to get the corresponding distance from the left hand side. Thanks everyone for watching. I hope this video helped see you in the next one.

Two people carry a heavy electric motor by placing it on a light board 2.00 m long. One person lifts at one end with a force of 400 N, and the other lifts the opposite end with a force of 600 N. (a) What is the weight of the motor, and where along the board is its center of gravity located?

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