A football coach sits on a sled while two of his players build their strength by dragging the sled across the field with ropes. The friction force on the sled is 1000 N, the players have equal pulls, and the angle between the two ropes is 20 degrees. How hard must each player pull to drag the coach at a steady 2.0\m/s?

Question

Diagram showing a box with friction force and two tension forces at 30 degrees.

Accepted Answer

Hey, everyone in this video, we're gonna be working through the following problem At a Carnival. A challenge is set up for two people to win the challenge. They must pull a rectangular box to the finish line, but both of them should have equal pulse. The angle between the two ropes tied to the box is 30°. If the friction force on the box is 970 Newtons determine how much force Each person must use to pull the box at a constant three m/s. Now, we're given a diagram of the situation. So we have our rectangular box. OK? And there is a friction force between the box and the surface of the ground. We have the two people pulling the ropes. OK? The first one has tension T one, the second one has a 10 T two and the angle between those two ropes is degrees and they're trying to pull it across the finish line. We're given four answer choices. Option a 222.7 nodes. Option B 382.6 Newtons. Option C 502.1 Newtons And option D 402.4 Newtons. Now we're looking to determine forces. So let's draw a free body diagram of our box. We have our box, OK. It's on the ground. So we know that we have the normal force acting up perpendicular to the ground. We have the force of gravity acting downwards on our box. We have the tension T one acting up into the right, the tension two acting down into the right and we have friction between the box. So the friction is going to oppose the motion. And so it's going to be pointing to the left. Now, this box is moving. So we know that we're gonna be dealing with kinetic friction F K. Now we're gonna take up into the right to be positive and we can break our attention T one and T two into their X and Y components. Now, we know the angle between T one and T two is 30 degrees and we want them to have equal pulse. And that means that the angle between each of them and the horizontal has to be the same. And so that angle is going to be 15 degrees. OK? Half of the 30 degrees that separates them. So we have our tension T1 in the X direction, our tension T two in the X direction and both of these or 15 degrees To tension one and tension two respectively. And then we have our tensions in the wide direction, the Y component of tension, one points in the upwards direction, The y component of tension two points downwards. OK. So we're looking for forces now recall that the, some of the forces, OK? And in this case, in either direction is equal to the mass multiplied by the acceleration. Hey, this is Newton's second law. It relates the forces to the mass and the acceleration. Now we're moving the block in the X direction. So let's look in the X direction first. OK. So we're gonna consider the sum of the forces in the X direction which is equal to the mass multiplied by the acceleration in the X direction. Now, in the X direction, which forces do we have, we have tension T one, X, tension T two X, the X components of the tensions acting in the positive X direction. So we have T one X Plus T two x and in the negative extraction, we have the force of kinetic friction. So we subtract that's OK. And this is equal to the mass multiplied by the acceleration in the X direction. Now let's see if we can break this up and write out as much information as we know and we don't know the tension T one, but we do know that T one X, OK is gonna be related to T one through the cosine because we're talking about the adjacent angle. So T one X is going to be T one multiplied by cosine of the angle. 15 degrees. T two, same thing. OK. It's also gonna be related through cosine because it's the adjacent side, we have the same angle 15 degrees. And so T two X is going to be T two multiplied by cosine of 15 degrees. Now, we're told that the force of friction is 970 Newtons. So we can substitute that in, we subtract 970 newtons and all of this is equal to the mass of the box multiplied by zero. Why am I multiplying by zero? Well, we're told that we want to move the box at a constant speed of three m per second. If we have a constant speed here, constant velocity. That means the acceleration is zero. The box is not accelerating. So acceleration is zero and the entire right hand side is gonna go to zero. Now, what else can we simplify? We're told that we want the two poles to be equal. OK. That means that T one Must equal T two and we're gonna call both of them key to simplify. OK. So these two tensions have to be the same. And so we can write T cosine 15 degrees plus T cosine 15 degrees -970 Newtons Is equal to zero. Yeah, we can simplify, this is equivalent to two T multiplied by cos degrees and that is equal to 970 newtons. And we're trying to solve for T. So now we can divide by two, multiplied by cosine 15 degrees. And we get that the tension T is equal to .109 Newtons. So we found a tension T of 502.109 nodes. OK. And remember that that is equal to both tension T one and the tension T two because we want both people pulling with the same force. So we found the force that each person must use is 502.1 newtons which corresponds with answer choice. C thanks everyone for watching. I hope this video helped see you in the next one.

Verified Solution

Key Concepts

Friction Force

Tension in Ropes

Equilibrium of Forces