A force of 176 lb makes an angle of 78° 50′ with a second forc...

Question

A force of 176 lb makes an angle of 78° 50′ with a second force. The resultant of the two forces makes an angle of 41° 10′ with the first force. Find the magnitudes of the second force and of the resultant.

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Accepted Answer

Hello to day, we are going to be solving the following word problem. So we are told that we have two forces A and B which are shown to be acting on a certain point. The angle between the forces A and B is 63 degrees with 40 minutes. We want to determine the magnitude of the force B and the resultant magnitude. If the resultant makes an angle of 27 degrees with 20 minutes with the force A, we are also told that the magnitude of force A is £812. So how can we go about solving for the magnitude of B and the magnitude of the resulting vector? Well, one thing to note is that we are allowed to move around vectors as long as we keep the same direction and magnitude. So if we move the vector A to be on top of vector B and the resulting vector, we can make an upper triangle. So just to draw this image, we have the vector B that is given to us and we have the vector A which has a magnitude of £812. And we have the resulting vector between these two vectors. Now keeping the same direction and same magnitude I can draw vector A to be above the vectors B and vector R and this vector will have the same magnitude of 812 degrees. We're going to use this triangle to solve for the magnitude of B and the magnitude of R. Now, what we need to do is we need to label the respective angles. Recall that the angle between A and the resulting vector is 27 degrees with 20 minutes with respect to that angle, the equal opposite angle will be the angle respective to the side B of the triangle. So we can label this angle here to be 27 degrees with 20 minutes. Now, we need to solve for the angle with respect to the side A. Now recall that the angle between the vector A and the vector B is 63 degrees with 40 minutes. But we also know that the angle between A and the resulting vector is 27 degrees with 20 minutes. So if we were to subtract the angle between A and B from the angle between A and the R and resulting vector, we will get the angle for B or the angle with respect to the side A. So we're going to do 63 degrees with 40 minutes minus 27 degrees with 20 minutes. This is going to give us a value of 36 degrees with 20 minutes. So this is going to be the value of the angle with respect to the side. A finally, we need to find the angle with respect to the R side of the triangle. So recall that the angle sum property for a triangle states that the sum of all three sides of a triangle should equal to 100 and 80 degrees. So if we take the two angles, we know which is 36 degrees with 20 minutes and sum it with 27 degrees with 20 minutes. And sum the angle with respect to R, this should equal 180 degrees. We can use this equation to sulfur the angle R. So 36 degrees of 20 minutes plus 27 degrees with 20 minutes will give us a value of 63 degrees with 40 minutes. We will now have 63 degrees with 40 minutes plus the angle R is equal to 180 degrees. If we subtract 63 degrees of 40 minutes to both sides of the angles or both sides of the equation, we will have an angle R is equal to 116 degrees with 20 minutes. This is going to be the length of the angle with respect to the R side of the triangle. So now we have all three angles and we have one at least one side of the triangle. How can we continue to solve for side B and the side R. Well, in order to do this, we're going to use the Law of Science, the law of Sciences states that we have three equivalent ratios, we have sign of angle A divided by S length A is equal to sine of angle B divided divided by the side length B equal to sine of angle R divided by the side length R. So what we're going to do is we're going to take two of these three ratios, set them equal to each other. And use that to solve for B. And for R, if we take a look at the triangle, we have the side length or the magnitude for A, it's 812. And we also know the respective angle to A is 36 degrees with 20 minutes. So we're going to take the ratio with respect to aim and set it equal to the ratio with respect to B in order to solve for the side length B. So let's go ahead and start by solving for the silent be or in other words, we are going to solve for the magnitude of B, we have sign of A divided by the side length A is equal to sine of B divided by the side length B. We will go ahead and solve for the side length B by cross multiplying these two values. This is going to give us S length B multiplied by sine of A is equal to S length A multiplied by sine of B. We will divide both sides by the sine of A. Leaving us with sine length B is equal to si length. A multiplied by sine of angle B divided by sine of angle A. Now recall that the side length of A is £812. The angle with respect to A is 36 degrees of 20 minutes and the angle with respect to B is 27 degrees with 20 minutes. Plugging in these values will give us B is equal to 812 multiplied by sign of 27 degrees or 20 minutes divided by sign of 36 degrees with 20 minutes. At this point, I would recommend plugging these values into a calculator. And when doing so, you should get a magnitude for B equal to £629.3. This is going to be the magnitude of the four speed. Now, we need to calculate the magnitude of the resulting vector. We're going to do the same thing as we did to solve for B. So now since we want to solve for the magnitude of the resulting vector, we're going to use the ratio with respect to A and set it equal to the ratio with respect to R. So now we are going to solve for R. So we have sign of A divided by sine length, A equal to sine of angle R divided by the magnitude R just like before we're going to cross multiply the two values and solve for R that is going to give us the side length R multiplied by sine of angle A equal to side length. A multiplied by sine of angle R. We will divide both sides of the equation by sine of angle A. And that will leave us with the side length R is equal to the side length. A multiplied by sine of angle R divided by sine of angle A. Now again, the side length of A is £812. The angle with respect to A is 36 degrees of 20 minutes. And we know that the angle with respect to R is 116 degrees with 20 minutes. So plugging in these values will give us the magnitude of R is equal to 812 multiplied by sine of 116 degrees with 20 minutes divided by sign of 36 degrees with 20 minutes. After plugging this value into a calculator, you should end up with a magnitude of R equal to £1228.3. This is going to be the magnitude of the resulting force between the mag, the vectors A and the vector B. So we figured out that the magnitude of B is equal to £629.3. And the magnitude of R is £1228.3. With that being said, the answer to this problem is going to be a. So I hope this video helps you in understanding how to approach this problem. And I'll go ahead and see you all in the next video.

A force of 176 lb makes an angle of 78° 50′ with a second force. The resultant of the two forces makes an angle of 41° 10′ with the first force. Find the magnitudes of the second force and of the resultant.

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Key Concepts

Vector Addition of Forces

Law of Cosines

Law of Sines

Watch next

A force of 176 lb makes an angle of 78° 50′ with a second force. The resultant of the two forces makes an angle of 41° 10′ with the first force. Find the magnitudes of the second force and of the resultant.<IMAGE>

Key Concepts

Vector Addition of Forces

Law of Cosines

Law of Sines

Watch next

A force of 176 lb makes an angle of 78° 50′ with a second force. The resultant of the two forces makes an angle of 41° 10′ with the first force. Find the magnitudes of the second force and of the resultant.

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