Suppose that you can lift no more than 650 N (around 150 lb) unaided.

(a) How much can you lift using a 1.4-m-long wheelbarrow that weighs 80.0 N and whose center of gravity is 0.50 m from the center of the wheel (Fig. E11.16)? The cen-ter of gravity of the load car-ried in the wheelbarrow is also 0.50 m from the center of the wheel. (b) Where does the force come from to enable you to lift more than 650 N using the wheelbarrow?

Video transcript

Hey, everyone in this problem, we have a male person that has a safe limit to lift objects weighing 250 Nunes K or 25 kg at the workplace were asked how much load can the person lift if they're using a 1.5 m long trolley weighing 100 newtons. The center of mass of the trolley is located 0.6 m from the axle supporting the wheels and were told to take the center of gravity of the load to also be located at 0.6 m from the axle. Alright, so we have our little diagram of our trolley here. We're gonna have our person who is lifting that trolley. When they lift the trolley, they're gonna exert an upward force. F okay. All right. And we know the maximum value that f can be okay, have a safe limit to lift objects weighing 250 Nunes. So the maximum value of F is gonna be 250 Nunes. Alright, then we have our load in our wheelbarrow. It's going to be right at the center of gravity of 0.6 m and it's going to have a downward force k of the weight, the weight of the load, it's going to be acting downward. Okay. And this is what we're trying to find. We're trying to find how much load can this person lift? What is that value of W L? Okay. Now, let's think about the other forces. Okay. We also have the weight of the trolley W T K acting downwards again from that center of mass and we have one more force. Okay. The wheel here is making contact with the ground. And so we're going to get a force on that wheel because of the ground and we're gonna call it F W force on the wheel. Okay. Alright. So, and let me maybe draw this in a different color because it's hard to see on the black wheel there. Okay. So let's take our coordinate system, okay. We're gonna take up into the right as positive, we have our X and Y directions. Now this is an equilibrium situation, okay. We have no net external forces acting on the system which tells us that the sum of the forces is going to be equal to zero and we have no net external torque acting on the system which tells us that the some of the torques is also going to be equal to zero. Okay. Now, when we have some of the force is equal to zero, this actually gives us two equations. Okay. We can break this down. Into our components. So we get that the sum of the forces in the X direction is equal to zero and that the sum of the forces in the Y direction is equal to zero. Okay. So we have three equations. So now we need to figure out which one to use. All right. So the first one, the sum of the forces in the X direction is zero, we actually have no forces acting in the X direction. Okay. So we're not going to be using that equation, the sum of the forces in the Y direction equal zero. Well, which forces do we have in the Y direction? We have F F W W L and W T. Okay. We don't know F W, we don't know the force uh the ground is acting on the wheel and we don't know the weight W L. That's what we're trying to find. Okay. So if we use the sum of the forces in the Y direction, we're gonna end up with two unknowns and we won't be able to solve. So let's go ahead and use the, some of the torques. Okay. When we do torque, we get to choose our pivot. Okay. If we choose our pivot to be the axle of this wheel, okay, then the force acting on this right by this axle of the wheel is not going to contribute to the torque. Okay. So it's not going to be in our equation. The only unknown in our equation is going to be W L and we're gonna be able to solve for that weight. Okay. So let's go ahead with the, some of the torques is equal to zero. Okay. And again, we're gonna take our pivot to be through the wheel access. Alright. So which torques do we have? Okay. What do we have to consider? Well, if we think about this upward force F, okay, it's going to be rotating. If you think about pushing up on F, you're gonna end up rotating around this axis counterclockwise. Okay. So that's gonna give us a positive torque. So we get the torque from F is positive. Okay. And then the torque from these weights, okay. They're pushing down that's going to cause clockwise rotation. So those are going to be negative. Okay. So we're going to get negative the torque of the trolley, negative torque from the load L okay. And these are equal to zero. Alright. So let's recall for torque, we have our F sine of the angle, okay. Where R is the distance from our axis F is the force and then sine sine of the angle. We're using the angle between the axis and our force. Okay. Alright. So we got our the trolley, the weight of the trolley is a force acting there times sine of the angle, theta. Okay. And same thing here and we'll call this theta t our troops and that's I did the trolley first we should have done the force at first. Sorry about that. Okay. So starting over are for the force, the radius for the force a times the magnitude of that force which is F times sine of the angle theta for that force. Now we have for our trolley, our trolley, the weight of the trolley signed data T and finally RL wait W L signed data L zero. OK. So let's just fill in what we know here. Okay. And we can imagine when the person first picks up this trolley, okay that the handle is going to be horizontal like this. Okay. This is where we start from. And so the angle that all of these forces is going to make with this axis is going to be 90°. Okay because that's our starting point for the rotation. Alright. So filling in the information R F the distance from the axle to our force F. Okay. Well, we know that this trolley is 1.5 m, okay. So that distance is going to be 1.5 m. So we get 1.5 m times f times sine of 90° minus. Okay. Now the distance from the axis or the axle, sorry to our center of mass of the trolley, which is also the center of mass of the load is 0.6 m. Okay. So for each of the next two torques are are is going to be 0.6 m 0.6 m times W T sign of 90 -0. m WL sign of 90. Alright, so we can divide by sine of 90°. Those will go away and we're going to end up with. Oops, I'm sorry we should have filled in the weight of the Raleigh we can do that next. Alright. Filling in our forces, the force F We said before that we have a maximum force f of 250 newtons. Okay. That's as high of force that this person can lift. So we're gonna southern f is equal to 250 newtons. Our weight of the trolley W T we're told Is 100 Newtons. Mhm. We have the weight of the trolley 100 newtons And this is all going to be equal to 0.6 m in terms of weight of the load which is what we are looking for. Okay. Alright and we've used the maximum force F that this person can lift which means that we're going to get the maximum weight of the load out of this. Okay? Alright. So on the left hand side, we get 300 and 75 newtons times meter minus 0.6 times 100 gives us 60 Newton meter Is equal to 0.6 m times the weight of the load. So if we divide by 0.6, we get that the weight of the load is going to be 315 Newton m divided by 0.6 m. The unit of meter will divide out. We're left with Newtons, which is what we want for a weight and we got newtons. Alright, so let's go back up and look at our answer choices and we found that the maximum load that this person can lift using this trolley Is going to be a newtons. Thanks everyone for watching. I hope this video helped see you in the next one.