A 1.0 kg wood block is pressed against a vertical wood wall by the 12 N force shown in FIGURE P6.57. If the block is initially at rest, will it move upward, move downward, or stay at rest?

Video transcript

Hi, everyone in this practice problem, we are being asked to determine whether a block or a box will move vertically up down or remain at its place. So as shown in the figure given in the problem statement, a force of 50 Newton is acting on a 2.5 kg steel box against a first lubricated steel surface. And the force itself is making an angle of 25 degrees with the horizontal axis. We're being asked to determine whether the block will move vertically up or down or remain at its place if it is initially at breast. And the options given are a slide upwards. B slide downwards C remain steady and D cannot be set. So we will model the box as a particle. And therefore, let's start with creating a free body diagram of our system where I am going to first draw the horizontal and vertical axis here. And let's just put our box essentially in the exact middle as a point here. So first I am going to depict the uh F bush or the Newton force acting on the box creating an angle of 25 degrees with the horizontal axis here. So this is going to be F Bush and F Bush is going to be equal to 15 Newton. So other than F bus, there are going to be other forces acting on the box. The first one is going to be obviously the weight of the box itself or F gravitational. So FG which is just essentially M multiplied by G and then the other force is going to be the normal force that is acting from the um lubricated steel surface onto the box. So that will be pointing towards the left just like so and that will be just a normal end. So those are the three forces acting on the box. And in this case, we know that the box does not move horizontally. So the horizontal summation of all the forces is going to be zero or Sigma FX is going to be equal to zero. So zero will be equal to negative N which is the normal force plus we have to actually project our F bush into the horizontal and vertical axis. So let's do that right now. So in the horizontal axis, it will be F push multiplied by cosine of 25 degrees. And in the vertical axis, it will be F bus multiplied by sine of 25 degrees just like so awesome. So now going back to our Sigma FX equals to 00 will be equal to negative N plus the F bush multiplied by cosine 25 degrees or the, um, X projection of F Bush multiplied by cosine of 25 degrees. And then, um, this will uh rearranging this, we will then get N to then be equal to just F Bush multiplied by cosine of 25 degrees. So now we can actually substitute our information that is given in the problem statement into this equation or expression that we have to calculate the normal force. So F Bush is going to be 15 Newton given in the problem statement multiplied by cosine of 25 degrees and calculating this, our normal force or N will then come out to a value of 13.59 Newton. Awesome. So now we wanna move on to the Y axis. So we wanna calculate Sigma Fy which will then be equal to F push multiplied by sine of 25 degrees minus FG or um substitute FG with M multiplied by G. This will then be equal to F Bush multiplied by sine of 25 degrees minus M multiplied by G. So essentially the goal of this step right now is to find which way the box is going to be moving into. So looking at this, we want to substitute all of our values that we know to calculate for Sigma Fy. So then Sigma Fy will then then be equal to F Bush which is 50 Newton multiplied by a sign of 25 degrees minus M which is given to be 2.5 kg multiplied by G which is 9.81 m per second squared. Calculating this, we will get Sigma Fy to then be equal to negative 18.16 Newton. So therefore, we know that the three forces in the free body diagram that we just uh did earlier will add up to negative 18.16 J Newton. So the box will remain ST steady if the force of static friction is plus 18.16 J Newton. So using the first equation that we have, which is the Sigma FX equals to zero, we want to actually determine the maximum static frictional force that we have. So, Fs Max will equal to us multiplied by N. And according to our calculation earlier, this can be calculated when us is given it to be 0.1 for lubricated steel surface. So let's use that value 0.1 multiplied by N which is 13.59 Newton. So calculating the uh maximum static frictional force, uh Fs Max will then come out to a value of 1. Newton. So the static friction force required to keep the box steady is more than the maximum value of the static frictional force. Hence, the box will actually slide downwards. So since FSX is a lot lower than uh the Sigma Fy negative 80.60 Newton, then the box itself is going to slide downwards. Therefore, the correct option is going to correspond to option B in our answer choices. So option B will be the answer to this particular practice problem and that will be it for this video. If you guys still have any sort of confusion, please feel free to check out our adolescent videos on similar topics available on our website. But other than that, that will be it for this video. Thank you.