Two forces 390 n at 12 and 311 n at 32 are applied to a car...

Two forces, 390 N at 12◦ and 311 N at 32◦ are applied to a car in an effort to accelerate it. 12◦ 32◦ What is the magnitude of the resultant of these two forces? Answer in units of N.

Transcript

0:00 Hi there is a question.

00:01 Say that forces are applied to the car like the given like as given in the field and f1 magnitudes are being given assuming assume up and to the right are in positive direction.

00:21 Okay, so what is the resultant vector of these two forces? i will be taking this as x this and y and suppose forces are applied here and this is is f2 30 degree in this direction this is f1 and 10 degree this is the origin so resultant force resultant force is i will be using the method of just resolving the vector or component form i will be writing vectors in component form and then we'll talk about let us say result it is this so f will be equal to f1 plus f2 in vector form so f1 plus f2 f1 will be it has f1 has two components one is along the y -axis another is in negative x -axis so it will be 10 sign 10 degrees i minus i because negative xx and 10 cost 10 degree j for f2 it is okay magnitude it is sorry for this this must be the magnitude of f1 being given as a 405 newtons four and five newtons plus 353 cost or in this case sign 30 degree high plus 353 cost 30 degree j so if you collect the coefficients it will be minus 405 sign 10 degree plus i'm collecting coefficients of i first 353 sine 30 degree i plus 405 cost 10 degree plus 353 cost 30 degree j so f vector i will be using calculator 353 sine 30 degree sine 30 degree minus 40 5 5 5 5 5 sign 30 degree minus 40 5 5 sign 10 degree which is 106 .1725 i plus 405 cost 10 degree plus 353 cost 30 degree which is 704 .5541 j cap.

03:26 So its magnitude simply f will be square of x component 106 .1725 whole square plus 704 .551 whole square under root so whole square plus 106 .1 1725 whole square under the square root so 712 .50 9090 okay so this is the magnitude newton's and to get the direction let us assume this to be theta so 10 theta will be y component divided by x component 704 .551 divide by 106 .1725.

04:39 704 .551 divided by 106 .17251 divide by 106 .17255.

04:50 So 6 .6359.

04:54 So theta will be 10 inverse 6 .6359.

05:02 So 81 .4303...

Question

Two forces, 390 N at 12◦ and 311 N at 32◦ are applied to a car in an effort to accelerate it. 12◦ 32◦ What is the magnitude of the resultant of these two forces? Answer in units of N.

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