The object shown in FIGURE EX12.29 is in equilibrium. What are...

Question

The object shown in FIGURE EX12.29 is in equilibrium. What are the magnitudes of $F sub 1 right arrow sub$ and $F sub 2 right arrow sub$ ?

Accepted Answer

Hi, everyone in this practice problem, we're being asked to find the magnitude of forces F one and F two acting upon a long thin wooden slap, which is going to be five m in length. The slap is of negligible mass and is in equilibrium. The forces acting on it are shown in the figure below. We will have first a 50 Newton force going upwards located right at the very edge or at the very left edge of the slap and then three m right off the 50 Newton force. There will be F one pointing downwards for acting on the wooden slap and then two m the right of F one, there will be F two also acting downwards to the wooden slap. The options given for the magnitudes of F one and F two are A F one equals negative 100 and 13 Newton and F two equals 75. Newton B F one equals negative Newton and F two equals 58 Newton C F one equals 134 Newton and F two equals negative Newton. And lastly D F one equals 125 Newton and F two equals negative 75 Newton. So we know that the slab is an equilibrium based on the problem statement. So when a slap or when a system is located at or S N equilibrium, then we can utilize FNET equals to zero, which is going to be translational equilibrium condition. And also Sigma torque equals to zero, which is going to be the rotational equilibrium conditions. First, what we wanna do is to apply the condition of the translational equilibrium since all the forces are acting along the Y axis, so all the forces are acting upon the Y axis, which means that Sigma F Y is equals to zero. So at the equilibrium, we know Sigma F Y equals to zero. So that means all the forces going upwards, which is in this case is going to be the 50 Newton di uh minus by all the forces going downwards which is F one and also minus F two, all of that will equals to zero. So F one in this case will then equals to 50 newton minus F two. I will call this our first equation which we will utilize later on when we have applied the second equilibrium condition, which is the rotational equilibrium. So applying the condition of the rotational equilibrium, we can take the torques around the very left corner of the wooden slap which in this case will allow us to make things a lot simpler because then the newton force can just be neglected. So OK, so Sigma torque zero, I would say at 50 newton, which in this case, we're looking at this point specifically right here. So the Sigma torque at this point or at this rotational axis will then equals to, we wanna recall that torque will equals to the force multiplied by the distance. So in this case, the first one that we had, we wanna look at is F one. So torque F one multiplied by the distance which is going to be three m F one is acting downwards. So that means it will be acting clockwise. So based on the convention of our torques, everything acting clockwise will have the negative sign in front of it or negative value. And then we wanna plus that with F two, which in this case is also pointing downwards, which means that the torque is going to be um clockwise in the clockwise direction. So I'm gonna add the negative sign. So negative F two multiplied by the distance from F two to the axis of rotation is going to be two m plus three m just like. So and all of that will actually equals to zero. So simplify this, we will then have negative three F one, negative five F two equals to zero, which means that negative three F one equals five F two and then rearranging this so that we have a, an equation for F one, F one will then equals to negative 5/3 of F two, which will be our second equation. So in this case, we have two equations with two unknowns which is equation one and equation two and two unknowns are F one and F two. Therefore, we can actually substitute either one of this equation into the other in order for us to actually solve what F one and F two is. So let's substitute equation two into equation one. So that means in equation one, we have F one equals 50 Newton minus F two. And we want to substitute our F one with our equation two. So that means negative 5/3 of F two will equals to 50 Newton minus F two. We can actually put the minus F two into the other side. So that will actually give us negative 5/3 of F two plus 3/3 of F two, which in this case is just F two equals 50 Newton. From here, we will have negative 2/3 or F of F two equals to 50 Newton. And after rearrangement or rearranging this, then we will have our F value being negative 75 Newton. After we found our F two value, we want to substitute that back into equation one. In order for us to actually find what F one is. So F one based on equation one, F one equals to 50 ne minus F two. And that will be F one equals 50 minus negative 75 which will come out to be F one equals 125 Newton just like. So, so there we go, we've found what our F one value is, which is 125 Newton. And we've also found what F two value is, which is negative 75 Newton. So that will actually correspond to option D in our answer choices. So option D will be the answer to this particular practice problem with an F one value of 125 Newton and F two value of negative 75 Newton. So that will be it for this particular practice problem. If you guys still have any sort of confusion, please make sure to check out our other lesson videos on similar topics and that'll be it for this one. Thank you.

The object shown in FIGURE EX12.29 is in equilibrium. What are the magnitudes of $F sub 1 right arrow sub$ and $F sub 2 right arrow sub$ ?

Key Concepts

Equilibrium

Torque

Force Components

Watch next

The object shown in FIGURE EX12.29 is in equilibrium. What are the magnitudes of F1→\overrightarrow{F_1}_{} and F2→\overrightarrow{F_2}_{}?

Key Concepts

Equilibrium

Torque

Force Components

Watch next

The object shown in FIGURE EX12.29 is in equilibrium. What are the magnitudes of $F sub 1 right arrow sub$ and $F sub 2 right arrow sub$ ?