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If the tension in rope 1 is 150 N, what is the tension in rope 2?

The sled dog in figure drags sleds A and B across the snow. The coefficient of friction between the sleds and the snow is 0.10. A is 100kg and B is 80 kg The figure is a dog ties to rope 2 that is ties to 80kg sled that is tied to rope 1 that is tied to 100kg sled

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1. Begin by drawing a FBD (free body diagram). I'm assuming sled A is behind the dog, and sled B is following sled A. Also assuming, sled A & B are being pulled by ropes 1 & 2, respectively. So, you've got a picture with the dog pulling two sleds. Draw an arrow pointing toward the dog from sled A with magnitude 150 N. This arrow represents the tension in rope A. Draw two arrows, one pointing downwards from each sled. These represent the weight of the sleds. The weight of sled A is equal to 100 kg * 9.81 m/s^2 (gravity on Earth). The weight of sled B is equal to 80 kg *9.81 m/s^2. We must also assume the dog is not accelerating. That is, the force pulling the sleds is not changing. OK, we've got our FBD drawn. Now, we must solve for the friction forces. The friction force always acts in the opposite direction to movement (friction opposes movement; it is dissipation of energy). So, draw an arrow on each sled pointing in the opposite direction from the dog. Our C.O.F. (coefficient of friction) is 0.1 for both sleds, so just multiply the weight of each sled by 0.1 and label each friction force arrow with its respective magnitude. Friction due to sled A = 0.1*9.81*100 = 98.1 N Friction due to sled B = 0.1*9.81*80 = 78.48 N From the above friction forces we now know that the sled is not moving at all. The total friction force (177.29 N) is greater than the force provided by the dog (150 N). However, we can now solve for the tension in rope 2. 150 N - 98.1 N = 51.9 N Tension in rope 2. This says that sled A is ready to move, however, sled B requires additional force to overcome the friction force that is holding it in place. If the sled were sliding, the tension in rope 1 would be equal to the sum of the friction forces, and the tension in rope 2 would be equal to the friction force due to sled B. After rereading your description it is clear that I've got my A's and B's and 1's and 2's backwards, that's just nomenclature. Hopefully this helps though! OK, since your description edit, this is what I understand: DOG ----rope 2-----[sled B]-----rope 1-----[sled A] Mass of sled B = 80 kg Mass of sled A = 100 kg Tension in rope 1 = 150 N Same deal as before. Draw an FBD. Friction force due to sled B = 80*9.81*0.1 = 78.48 N Friction force due to sled A = 100*9.81*0.1 = 98.1 N same as before to this point The difference here is that the tension given is in the rope between the two sleds, and not between the dog and the sled. Represent the tension as an arrow "pulling" (pointing away from the dog) on sled B. This problem gets a little funky here. Sled A must be accelerating, unless there is some other force besides friction acting on it. We'll assume there isn't. F=ma, M = 100 kg, F = 150 N - 98.1 N = 51.9 N So, a = 1.93 m/s^2 kinda weird. Again, F=ma (this time for sled B) We're assuming the sleds accelerate at the same speed. Otherwise the rope would break, or somehow sled A would run into sled B or sled B would run into the dog. We're also assuming the acceleration is positive as it doesn't specify that there are breakes on the sleds that would provide a force other than friction to oppose the motion of the sled. So, a = 1.93 m/s^2 -------> F = ma -------> F = 80*1.93 = 154.4 N This 154.4 N is the force required to accelerate that mass at 1.93 m/s^2. In addition to this force, we must also balance the friction force due to sled B, and the tension in rope 1 pulling on sled B. 154.4 N + 150 N + .1*80*9.81 = 382.9 N Final Answer: Tension in rope 2 = 382.9 N
2. where is the figure???
3. Let: Ft = force applied on the tow rope F = inertial force of the sledge Ff = Frition force m = mass a = acceleration Ft = Ff + F F =ma For rope 1: Ft = 150N, Ff = 100x.1 = 10N; Weight = 100kg 150 = 10 +(100/9.81) a a = 140/(100/9.81) For rope 2: Ff = 10 + 80 x .1 = 18N; Weight = 100 + 80 = 180 kg Ft = 18 + (180/9.81)(140/(100/9.81) = 18 + 180(140/100) = 270N
4. Mass of A = m1 = 100 kg Mass of B = m2 = 80 kg Coeff of kinetic friction between sleds and snow = µ = 0.10 Tension in rope 1 = T1 = 150 N Tension in rope 2 = T2 = ? Both sleds have same acceleration. Let acceleration = a Draw FBD of A. Forces acting on A are: - 1. Weight m1g downward 2. Normal force N1 from snow upward 3. T1 in direction of motion 4. Friction f1 opposite to direction of motion. There is no acceleration in vertical direction. Therefore, total force in vertical direction = 0 Or, N1 - m1g = 0 Or, N1 = m1g f1 = µm1g Total force in horizontal direction = T1 - f1 = T1 - µm1g Therefore, a = (T1 - µm1g)/m1 ---------(1) Likewise, draw FBD of B. You will find that a = (T2 - µm2g)/m2 -------(2) From (1) and (2) (T2 - µm2g)/m2 = (T1 - µm1g)/m1 Or, T2 = µm2g + m2 * (T1 - µm1g)/m1 = 0.10 * 80 * 9.8 + 80 * (100 - 0.10 * 100 * 9.8)/100 = 78.4 + 80 * (100 - 98)/100 = 78.4 + 1.6 = 80 N Ans: 80 N