The 1.0 kg block in FIGURE EX7.23 is tied to the wall with a r...

Question

The 1.0 kg block in FIGURE EX7.23 is tied to the wall with a rope. It sits on top of the 2.0 kg block. The lower block is pulled to the right with a tension force of 20 N. The coefficient of kinetic friction at both the lower and upper surfaces of the 2.0 kg block is μ_k = 0.40. What is the tension in the rope attached to the wall?

Accepted Answer

Hey, everyone in this problem, a block weighing 1.5 kg is placed on another block weighing 2.5 kg, the blocks are tied as shown in the figure. And when the figure we have our 2.5 kg block resting on the ground with the tension force of 25 newtons pointing to the left. And our 1.5 kg block is sitting on top of that 2.5 kg block with a cable rope tied to the wall on the right hand side. So we're told that tension force in the lower rope we talked about is 25 mutants and the coefficient of kinetic friction is 0.35 for both the surfaces of the lower block which are in contact. And we're asked to determine the tension force in the upper rope. We're given four answer choices all in Newton's option A 25 option B 15, option C 9.4 and option D 5.1. So we're looking for a force and so let's start by drawing some free body diagrams. So we're gonna stick with the colors we have here. So the upper block block we're gonna do in blue and we have our upper block and we have that walk and we want to think about the forces acted. Now, this is resting on a surface, OK. It's resting on the surface of that lower block. And so the box is gonna have a normal force pointing upwards. And we're gonna say that this is a normal force of the lower block acting on the upper block. NLU. We know there's also gonna be a force of gravity. So we have the force of gravity acting on that upper upper box FGU. We have this rope tying it to the wall. So we're gonna have a 10 force acting to the right and we know that there's friction between these blocks. OK? Now, because this bottom block is moving to the left, the friction force on our upper block is also going to be pointing to the left. I need to oppose the, we have this friction force between the lower and upper block flu. So that's our free body diagram for the upper block and we can do the same for the lower block for the lower block. There's gonna be a few more forces involved. So we're gonna start with the normal ones. We have the um force of gravity of our lower block acting downwards. We have the normal force acting on the lower block NL from the ground. OK. So that's just the block is resting on the ground. So it has some normal force, then we have this other force. OK. On the upper block, we had this force from the lower block acting on the upper block. So in our lower block, we have to have that action reaction pair, we're gonna have this normal force from the upper block acting on the OK. So that upper block is imparting some force by resting on top of that lower right, and that's pointing down. Now, our attention were shown in the diagram points to the left. So we have our attention force t of the lower block and we have some friction to deal with. OK. Now, the first friction is the regular friction. We would think about the friction of this lower block with the route, but we also have friction at the top with that, that upper block. And so again, we have this flu on the upper block pointing to the left. We're gonna have an action reaction pair on the right and we're gonna call this ful OK. So these are equal magnitude opposite in direction, action reaction pair forces acting between those two blocks. OK? We have all our forces drawn on our diagrams. We're gonna choose the direction of up into the right to be positive. And we're gonna start with the calculations. Now, we want to determine the tension force of the upper rope. So let's start with the upper block. OK? And that tension force is in the X direction. So we'll start in the X direction, the sum of the forces acting on the upper block in the X direction, it's gonna be equal to the mass of the upper block multiplied by the acceleration of the upper block in the X direction. OK. Newton's second block. Now the sum of the forces in the X direction. Well, we have the tension acting in the positive direction. T and we have this four student of friction, flu acting in the negative direction. So T minus flu is equal to the mass multiplied by the acceleration. But in this case, that's gonna be zero and that's gonna be zero because this block is tied to the wall. And we can see that it's tied with this rope. So it is not going to be moving horizontal. OK. All right. So our acceleration is zero that makes the entire right hand side zero. And what that tells us is that this tension that we're looking for T is just going to be equal to this force of friction flu. Now, this force of friction, we have to figure out if it's kinetic or if it's static. And there's a bit of a hint in the question, we're only given the coefficient of kinetic version, but we can also think about it that even though this top block is not moving, OK, the block underneath it is moving. OK. So there's some relative motion of this top block relative to that bottom block where the friction is occurring. So this is kinetic friction. OK. So we're gonna have that, that friction force is gonna be equal to UK the coefficient of kinetic friction multiplied by the normal force on this block. And that normal force is NLU. Now we know the coefficient of kinetic V but we need to find NLU in order to calculate our attention. So let's go back to our diagram. OK. We're sticking with the upper block and now we're gonna switch over to the, some of the forces on that upper block in the Y direction because that's where we're gonna see this NLU force. That's where we can calculate that normal force that we need to then calculate our friction and our 10. Now some of the forces in the Y direction is going to be zero. This block is not moving up or down, there's no acceleration in the Y direction. So the forces is equal to zero in the positive direction. We have NLU in the negative direction FGU. So NLU minus FGU is gonna be equal to zero. It tells us that the normal force we're interested in is just gonna be equal to the force of the grav force of gravity acting on that upper upper block. And we know that that's gonna be the mass of the upper block multiplied by the acceleration due to gravity. OK. So we have this NLU value that we're gonna substitute in. Now, to our tension equation. This tells us that the tension is gonna be that coefficient of kinetic friction in the UK multiplied by the mass of the upper block multiplied by the acceleration due to gravity. We know all three of these values we can substitute them in and solve for the tension. And we don't even need to use all of that information we were given about the lower block. OK. Everything we've done to this point has been dealing with that upper block and that is enough information for this problem. So let's substitute in our values and get this final answer. T is gonna be equal to 0.35 multiplied by 1.5 kg, multiplied by 9.8 m per second squared. And when we work this out on our calculator, we get 10 of about 5.145 newtons comparing this to our answer choices. We can see that this matches with answer choice D OK? And so the tension in that upper rope is gonna be 5.1 newtons. Thanks everyone for watching. I hope this video helped see you in the next one.

Key Concepts

Newton's Second Law of Motion

Frictional Force

Tension in a Rope

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