Two people are carrying a uniform wooden board that is 3.00 m long and weighs 160 N. If one person applies an upward force equal to 60 N at one end, at what point does the other person lift? Begin with a free-body diagram of the board.

Video transcript

Hey everyone. So today we're being given a problem about a systems equilibrium. So we're being told that a uniform pine plank of length, two m and a mass of 15 kg is being held by a carpenter and his son. The carpenter applies a vertical force of magnitude 40 newtons on the left end. And with this information, we're being asked to find one the magnitude and to the point of application of the force that the carpenter's son should exert to the main board or to maintain the board stable. E we're being asked to draw a free body diagram of the plank. So here we have the plank already. So let's actually just add on to what we have here where the carpenter is, well, where the carpenter's supporting the end is the axis, right? And we know that it's a length of two m and the weight of the board is coming down in the center of the board. So we know that this should be at this distance from at least from the carpenter to the center, Should be one m. We have the rest of the forces that are here. We have the weight of the board as well as F one. The force applied by the carpenter to support the plank as well as the carpenter's son, who is a indeterminate distance away from the axis and is also applying a force upwards to maintain the more stability. So we're going to need both conditions of equilibrium to go ahead and solve this. The first condition being that the force, the sum of all forces will be equal to zero. And the second condition saying that the sum of all torques and the system will be equal to zero. So let's take a look at the forces first and I'll draw this in blue oops. The sum of the forces of all forces will be zero, which means that this will be F one plus F two minus MG. Right? Because we have F one and F two that are upwards minus the force of weight of the board, which is simply mass times gravity. Now we know that this is equal to zero. So we can say that F two is therefore equal to M G minus F one. Because we're given F one, it's 40 newtons were also given that mass is 15 kg and we know that G the force of acceleration due to gravity is simply 9.81 m per second squared. So we can therefore say that F two is equal to 15 kg, 15 kg, Multiplied by 9.81 m/s squared -40 Newtons. Because again, F one is 40 newtons, it's 40 newtons. So F two. The force that the sun is applying, the magnitude of the force of the sun is applying must be 107.15 point newtons. Now we can go ahead and take a look at the second condition of equilibrium, which states that the sum of all torques must be equal to zero. So here We can use torque. So and F two times X minus the weight of the board, times the length from the carpenter to the Centre point is one this one m, which means if we want to find X, if we want to find the distance or the point of application from the start of the board all the way to where the force is being applied by the sun, which is X. We can rearrange and say that X is equal to W by F two and we know that the weight is again, MG. This is 15 Kg into nine point 81 m/s squared divided by and F two is, as we discovered earlier is 107. newtons. So finally we get that the distance that the sun must be has to be 1.37 m from the left from the left And he's applying a force of 107.15 newtons. If we look at our answer choices, we can see that this corresponds with answer choice. D. The sun is 1.37 m from the left end and applies a force of 107.15 newtons to maintain the board stability. I hope this helps. And I look forward to seeing you all in the next one