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41. Two ropes are attached to the bumper of a car One person pulls the first rope with a force of 50 pounds at an angle of 450 to the horizontal ground, while anoth...

Question

41. Two ropes are attached to the bumper of a car One person pulls the first rope with a force of 50 pounds at an angle of 450 to the horizontal ground, while another person pulls the second rope with a force of 70 pounds at an angle of 359 to the horizontal ground: The same effect can be produced by a single rope pulling with what force and at what angle to the ground?

41. Two ropes are attached to the bumper of a car One person pulls the first rope with a force of 50 pounds at an angle of 450 to the horizontal ground, while another person pulls the second rope with a force of 70 pounds at an angle of 359 to the horizontal ground: The same effect can be produced by a single rope pulling with what force and at what angle to the ground?



Answers

Two persons pull horizontally on ropes attached to a post, the angle between the ropes being $60^{\circ} .$ Person $\mathrm{A}$ pulls with a force of 150 lb, while person $B$ pulls with a force of $110 \mathrm{lb}$ (a) The resultant force is the vector sum of the two forces. Draw a figure to scale that graphically represents the three forces. (b) Using trigonometry, determine formulas for the vector components of the two forces in a conveniently chosen coordinate system. Perform the algebraic addition, and find the angle the resultant force makes with A.

Here we have the car being towed. We have the weight of the car acting straight down. We haven't. We have the car already on an incline of five degrees. We have the normal force acting perpendicular to the surface of contact. And we have this tension force acting at an angle of 10 degrees. Yeah, with respect to with respect to the with the with respect to the horizontal and so or the the true X access. And so you can see that here, that 10 degree where it would be essentially split by that axes by the dashed axes here and so we can first find for Part B. We first apply the Newton's second law, and we can say that then some of forces in the X direction would be equal to the mass times acceleration the extraction. This would be equal to zero because we want the car to be moving at a constant speed, which means that then the weight of the car sign of five degrees must be equal to the tension with which it's being pulled co sign of thought degrees. And so we find that then the tension is simply equal to the weight multiplied by tangent of five degrees. This would be equal to, uh, the mass of 1000 kilograms multiplied by 9.8 meters per second squared and this would be multiplied by tangent of five degrees. Therefore, the tension must be equal to approximately 860 Newtons. That would be your answer for part B. That is the end of the solution. Thank you for watching.

All right. So we have to force is here with an angle of 72 degrees between them. They're both equal forces. Someone's going to call them both X. And what we're really interested here is this half angle 36 degrees Because we know the why component of this force of these forces is going to be X times the coast of 36. So that's this force. Here will be X coasts 36 we have two of them. Cement two times X coasts of 36 equal what is given, which is 3 72 And if you solve for X, you will get a force of 2 29.9 rounded up to 2 30 Newtons.

Okay, Question 27. You pull a rope oriented at a 37 degree angle about the horizontal. The other end of the rope is a test. The front of the first of two wagons that have the same 30 kilogram mass on the roof exerts a force of magnitude. T bone attention, one on the first wagon. The wagons are connected by a second horizontal rope that exerts a force of magnitude teaching on the second wagon and determine the magnitudes of T one and T two if the acceleration of the wagons is two meters per second squared, so what we're looking at here is just two wagons connected by a group and then 1/3 a second rope pulling at an angle off 37 degrees. And this is T one on T two exists in between the two where, as we remember, tension always pulls so 30 kilograms in both cases for the masses and then we just need to work this hour using simple separate force diagrams. So we begin with two on one. We'll call these cases two on one based on which broke is putting them to the right. Two is easier from two, we can see that the only force acting that's of interest is the tension to cause. Any normal force of gravity is going to balance out, and 32 is simply going to be equal to the mass of the block multiplied by the acceleration, which is to and therefore attention to is equal to 60 Newtons. In the case of Warm, we can look at this slightly differently. That t one it's co sign of 37 is going to be pulling it to the right, whereas T two is going to be putting it to the left, and that's going to be equal to 30 multiplied by the acceleration as both blocks accelerated the same rich. That's basically because, he wrote, attaching them is not some kind of bungee cord on. So from this we can see the T one minus 71 co sign of 37 minus 60 is equal to 60 itself, so T one co signer, 37 is equal to 120 So fine t one. We simply divide 120 by the care sign of 37 and we pop this into a calculator giving us t one to be equal to 150.2. Newton's remembering that we correct for significant figures as we bought 150 is probably sufficient here. So that's 150.2 Newtons, or 150 Newtons for T one and 60 Newton's 42 and that's question.

Called the wagon wasn't diagram. So basically, this is the holy gentle service over there. Onda? Uh, yeah, Let me try it. Let me try it. Our gentle surface way to one ways. True. This is one. Let me draw it better person. Come on. Yeah, this is 12 on And this one and and actually, yeah, on the angle between the rope and the hard generally 37 degree and no friction over there. So a mass off both of the Martha, which is equals to that started K, G and Onda. We know that this force, there will be attention in this gesturing over there and which have two component Andi, obviously, when there will be m g force and normal Force MGI a normal onda Uh, this is for the war on two. So on the Tencent, which is pulling this one, is t t to let's say and obviously to to Tito in this when we also over there on the t one would be this one like this one said said C one So we're gonna exploration. We can see that for the second one. The explanation for the two in the extracts and would do nothing but just need force in the direction affects and then upon the mosque. So the net force over there Egypt, it's the net force in the direction effects on the to upon mass, which is equals to just 2 m per second Squire. And by this logic, we can say that the decision was this forces t two x or t two will be nothing but a mass times acceleration mass over there is 30. So 30 times of two will be 60 Newton and do. But this logic, we can say that and exploration of the work on one is also same. So and that exploration off the wagon one in the direction affects is nothing but than the original component off the one which is t one cost heater. The one cause off 37 degree minors off t two. So which will be equals toe the and this is equals to two. So basically on Duh Mm hmm. Actually upon to mass over there as well. So, basically is to so we can say that t one because 37 4 upon five minus t two, which is we have calculated 60 equals to mass stamp of two, which is 60 as well. So 30 in the sixties. So 60 61 2031 would be 1 20 into five upon four, which will be 30 30 over there. 4. 30 on the 1 15 Newton, okay?


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