A crate is supported by two ropes. One rope makes an angle of ...

Question

A crate is supported by two ropes. One rope makes an angle of 46° 20′ with the horizontal and has a tension of 89.6 lb on it. The other rope is horizontal. Find the weight of the crate and the tension in the horizontal rope.

Accepted Answer

Welcome back. Everyone in this problem. A trunk full of fruits and vegetables is tied by two strings in the opposite direction. The first string makes an angle of 38 degrees, 30 minutes with the ground and has a tension of £76.5. The other string is parallel to the ground. Calculate the weight of the trunk and the tension in the other string. For our answer choices. A says the weight of the trunk is £47.6 and tension is £59.9. B says it's £59.9 and £47.6. C says it's £67.8 and £35.7. And D says it's £35.7 and £68.7 respectively. Now, before we solve, let's try to sketch or visualize what's really going on here. OK. So we have a trunk full of fruits and vegetables. Let's say that uh our, our fruits and vegetables, our box of fruits and vegetables looks like that. OK? And for our trunk, it has a tension of £76.5. So we can draw that tension here. OK. And let's call that tea for attention and it has another string that's parallel to the ground going in the opposite direction. So let's say that it's heading this way, it's parallel to the ground in this case, or X axis. OK. And let's call this one T prime. And now we also know that our first string makes an angle of 38 degrees. OK? And 30 minutes with the ground, I think I should write that below it. It's quite a lot to write out. OK. And we also know that it has a tension of £76.5. Now, we're trying to figure out the weight of the trunk. OK. So we can represent the weight of the trunk here as w and we're also trying to figure out the tension in the other string. So we're trying to figure out T prime. So these are the two things that we want to know. Now, how can we figure it out? Well, notice here that our weight is acting vertically and the tension in the other string is acting horizontally. So they form a right angle degree with sorry, they form a right angle with each other. And if we think about the tension in our string that is angled at 38 degrees and 30 minutes, that means its vertical component is going to be the weight of the trunk are equivalent to the weight of the weight of the trunk and its horizontal component is going to be equal to the tension in the string that is parallel to the ground. So if we can find the vertical and horizontal components of the tension or of the first string, then we should be able to find the weight of the trunk and the tension in the other string. OK. So let's see if we can do that first. Let's start with the weight of the trunk. No, we have the anger, we have the anger of the uh we have the anger that is formed between the ground and the string, the se the the first string. OK? And the weight of the trunk is opposite to that anger. So let's put that in black here as opposite. Likewise, the tension in the string parallel to the ground would be adjacent to the anger that we know and the tension would actually be the hypotenuse. The value of the tension would be the hypotenuse of the right angle triangle that is found. Therefore to solve for the weight solving for W OK? Where we let w be the weight of our trunk, then if we think about it, we want to find the opposite side. We already know the high party news. Let's ask ourselves what trigonometric ratio links opposite and high party news. We can use this the sign ratio because recall that the san ratio tells us that the sign of an anger is the same as the ratio of its opposite side to its hypotenuse. So in this case, then the sign of our anger 38 degrees and 30 minutes is going to be equal to the opposite side, which is W divided by our hypotenuse, which is £76.5. Therefore, the weight of the trunk W is going to be equal to 76.5 multiplied by the sine of 38 degrees and 30 minutes. Now, when we put that into our calculator, we find that the weight of the trunk is going to be equal to £47.6. So that would be the weight of the trunk. The vertical component of our diagram. Now that we have solved for W we can also solve for the tension in the other string that's parallel in the ground that's T prime. No solving for T prime. OK. So going for T prime here, what do we notice again? We said that it's adjacent to the anger that we already know 38 degrees, 30 minutes. What trigonometric ratio links adjacent side to the high potter news? Well, it's the core side recall that the core sign of an angle is the ratio of its adjacent side to its hypotenuse. So this tells us then that the cosine of our angle 38 degrees and 30 minutes equals the adjacent side T prime divided by the hypotenuse which is 76.5 points which like we had figured out in our previous solution, that means T prime must be equal to 76.5 multiplied by the core time of 38 degrees and 30 minutes. Now, when we put that into our calculator, we find that the tension in the other string works out to be 59.9 points. Therefore, the tension in the string, the weight of the trunk is £47.6 and the tension in the string parallel to the ground is £59.9. If we look back on our answer choices, that would have been answer choice. A thanks a lot for watching everyone. I hope this video helped.

A crate is supported by two ropes. One rope makes an angle of 46° 20′ with the horizontal and has a tension of 89.6 lb on it. The other rope is horizontal. Find the weight of the crate and the tension in the horizontal rope.

Key Concepts

Resolving Forces into Components

Equilibrium of Forces

Trigonometric Functions and Angle Conversion

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