Three sleds are being pulled horizontally on frictionless hori...

Question

Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E $5.14$ ). The pull is of magnitude $190$ N. Find the tension in ropes $A$ and $B$ .

Three sleds of different weights (30 kg, 20 kg, 10 kg) are being pulled horizontally on ice with a force indicated.

Accepted Answer

Hey everybody. So today we're being asked to calculate the tension in each rope in a system where three boats are being pulled along the X axis via frictionless or on a frictionless surface. Now we're being told that the magnitude of the force of the pulling force being applied is 480 newtons. Now the first thing we need to do is consider the entire system as one singular component. And this is because the acceleration experienced by the each boat will be the same, right? The altering boats will have the same horizontal acceleration due to there being one, one singular force that is pulling all three of them. And we're not experiencing any friction forces. So we don't have to worry about that. Let me erase this and draw it a little better. But if we're to consider this as a excuse me to consider this as a composite as a composite system, then we have a uh box where we have, let's say we have a pulling force. We would have a normal force perpendicular to the surface of motion which is in the horizontal X direction. Right? And we also have the total weight, which is the mass total of all three boxes into the acceleration due to gravity or g Now for to solve that For to solve that. Well, we can use Newton's second law, which is that force is equal to mass into acceleration. We know what the force and mass are, which means acceleration is equal to force over mass, which is simply 480 newtons over while plus 20 plus 10 is equal to 60 kg. So that will be a total mass of 60 kg, giving us an acceleration to 8.0 m per second squared. And this acceleration is important because now we have to go back and find what are the tensions or the values for tension between each boat. So let's draw that out below. So to find the tensions and scrolling down a bit. So we have more space to draw. Let's redraw our diagram. So we have three boxes, right, Or three boats. Rather three boats. No, We have the first boat with a massive 30 kg. We have the second boat, which has a massive 20 kg. And the last boat with a massive 10 kg. We also have the pulling force P and we also have some tension forces. Now this is rope A. So we have a tension force here as well as here, it'll be acting on both of these boxes and it'll be going away from the center of the rope because it's being pulled on either side by the boats. And we have similar reactions going on over here for tensions in rope B. So this attention be this is tension B as well. And finally, each box will experience a a normal force because again, these uh these boats are on a horizontal surface on a horizontal surface, which means they'll also have a normal force that opposes the force due to gravity. So, to find the tension in uh or to find the value of tension in rope. A To start off with, we can say that using Uh the first boat, the 10 kg boat as an example. So we can apply Newton's first law. We're talking about force in the X direction, multiplied by the acceleration in the X direction. The sum of the forces on the 10 Kg block Or on the 10 KK boat. I keep thinking of them as blocks. The sum of the forces will be the force pulling it minus the tension in rope. A And that will be the mass of the 10 kg blocks, which is just 10 kg. Multiplied by acceleration, which means to find tension. Therefore the tension in rope A will therefore vehicle to p minus Block 10 into a substituting in our values, we get 480 newtons minus 10 kg into 8. m per second squared, Which gives us 4 80 80, which equals 400 newtons. We can do a similar thing for attention. B right, And we can use the mass, the 30 kg Block for that. Let's draw that in red. So following a similar plan in mind, we get in the X direction is equal to mass of the 30 kg block into acceleration in the X direction. The sum of the forces here. Well, It will simply be the only force acting here is the force to detention, which would be massive. 2 30 Or mass a kilogram or both. 30 into the acceleration, which can be simplified to 30 kg into 8.0 m/s squared Which gives us 240 newtons. Therefore our answer that we can select will be answer choice D. The tension in rope, a tension in rope A is 400 newtons and tension in roby is 240 newtons. I hope this helps. And I look forward to seeing you all in the next one.

Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E $5.14$ ). The pull is of magnitude $190$ N. Find the tension in ropes $A$ and $B$ .

Key Concepts

Newton's Second Law of Motion

Tension in Ropes

System of Connected Objects

Watch next

Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E5.145.14). The pull is of magnitude 190190 N. Find the tension in ropes AA and BB.

Key Concepts

Newton's Second Law of Motion

Tension in Ropes

System of Connected Objects

Watch next

Three sleds are being pulled horizontally on frictionless horizontal ice using horizontal ropes (Fig. E $5.14$ ). The pull is of magnitude $190$ N. Find the tension in ropes $A$ and $B$ .