Four forces are exerted on the object shown in FIGURE $P 3.45$ . (Forces are measured in newtons, abbreviated $N$ .) The net force on the object is ${\vec{F}}_{net} = {\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} + {\vec{F}}_{4} = 4.0 \hat{i} N$ . What are (a) $F ⃗ sub 3$ and (b) $F ⃗ sub 4$ ? Give your answers in component form.

Question

Four forces are exerted on the object shown in FIGURE $P 3.45$ . (Forces are measured in newtons, abbreviated $N$ .) The net force on the object is ${\vec{F}}_{net} = {\vec{F}}_{1} + {\vec{F}}_{2} + {\vec{F}}_{3} + {\vec{F}}_{4} = 4.0 \hat{i} N$ . What are (a) $F ⃗ sub 3$ and (b) $F ⃗ sub 4$ ? Give your answers in component form.

Accepted Answer

Hey everyone. So this problem is working with vectors. Let's see what they're asking us. You know that a toy has four forces acting on it. The unit of force used is newtons denoted by the letter N. The sum of the forces on the toy is sum of F equals FA plus F B plus F C plus F D equals six J newtons. So that's six newtons acting in the J or Y direction. We're asked to determine the values of F sub B and F sub C expressing the result using components. So in this graph here we are shown the magnitude and direction of all four of our forces. And then we're given multiple choice answers below. So we have a F of F of B equals negative 4.33 newtons in the I direction F of C equals negative 3.5 newtons in the J direction or choice B F sub B equals 4.33 newtons in the I direction. And F sub C equals 1.5 newtons in the J direction C F sub B equals negative 2.5 newtons in the eye direction. F sub C equals 3.33 newtons in the J direction. For choice of D F sub B equals 4.33 newtons in the I direction. And F sub C equals negative one newton and the J direction. OK. So they've asked us to report our answer using components. So that means breaking up the vector instead of a magnitude and an angle into its components in the X and Y or relatively I M J directions. So the first thing we're gonna do here is break up is is write out each of these four forces. So F of a, we see is a magnitude of seven newtons and it is in the positive Y direction. So we will write that as seven newtons check and there is no X or I component absent B the force, the magnitude of that is all in the X direction in the positive X direction. So that's gonna be the magnitude of F F B which we are going to solve for. And that's in the eye direction For the four c. We can see that that is in the J direction. So the magnitude of that course at sub C in the J direction and then F MD as both X and Y components, it's given to us as a magnitude of five newtons with an angle of 30 degrees below the negative X axis. So the way that we'll break that up into our X and Y components is by saying that in the X direction, we have minus five newtons times the cosine of 30 that was gonna be in the eye direction in the J direction. We also have -5 Newtons Times The sign of 30. And so it's minus, both of these terms are minus five or negative five because we are working in quadrant three where we have negative X components and negative Y components to make up that force factor. All right. So we can simplify this. We plug this into our calculators and we get -4. Newton's in the eye direction minus 2.5 s in the J direction. So that is hour at sub dating, we are told that the sum of the forces is equal to six newtons in the J direction because there is no I component. We can recognize that the I component is zero. And so we'll write that as FA X plus F B X plus F C X plus F D X equals zero, FA has no X component. So it's going to be zero F B sub X, we know has an X X component that is just the magnitude of F sub B RFC also does not have an X component. So that is zero. And then F D, our X component is negative 4.33 newtons. So I'll write that as minus 4. newtons equals zero. So you simplify that and see that FA B equals 4.33 newtons in the eye direction. So when we look at our multiple choice answers for this first part, we can eliminate choices A and C. Right now, we're going to do the same thing in the Y direction. So F A Y plus F sub B Y plus F sub C Y plus F sub D Y All equal six, the sum of that equals six. So when we take our Y components, power F sub a has a positive seven Newton Y component or FA B does not have a Y component. So that's zero. Our F sub C does have A Y component is the full magnitude of F sub C. And then our F of D has a component of negative 2. mute and not all Equals six new ones. So when we simplify that out and so for F sub C, we get 1.5 newtons in the J direction. And so here we see that, that aligns with answer choice B so B is the correct answer for this problem. That's all we have for this one. We'll see you in the next video.

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Key Concepts

Net Force

Vector Components

Trigonometric Functions in Force Analysis