Question

omauslysics/tecture/Lecture Homework 1 1. Your friend has slipped and fallen. To help her up, you pull with a force F, as the drawing shows. The vertical component of this force is 130 newtons, and the horizontal component is 150 newtons. Find a- the magnitude of F and b- the angle 0 2. Consider the vectors shown B (80.0 N) 30.0A (100.0 N) 30.0 53.00 C (40.0 N) a- Find the components of each vector (A, B, and C) b- Write each vector in terms (lineal combination) of unit vectors i and j c- Find the magnitude and direction of the vector D A-3B+ C 2 o- ume angie

can you explain how to solve for question C.

Answer 1

) (80 (oo N) first we have to find the component of à E, and 2 in x any y direction Ā = (100 Cos 30º) î + (100 sin 30") j) N à = (5053 it sos) Newtons/ 2140 N) B : (80 sin 309) (-) + 80 605 30° Ś * B = 640 î + 40 JZ ] New tous {sin 30" = { 1 Ĉ - (- 40 605 530) î + tosin 53• (-1) ) 17: 624€ - 325) News tous at Sapproximated Value of] | Sin 53" and los 53°S Į is taken {sin 53° -4/5 Now, we have to find the vector 3 6553° = 3/5 and 5 = 1/2 à - 38 t? à = ² (50531 2.50f) = 25857 +255 3B": 34-40 + 40V5j) = -120 î +12053ſ c = (-24î - 325) = (25w3i+ 253 )-(-120 i +120 335) + (-24 î - 323) Adding two vectors. - (2573 +120-24) { + (25-12083 -32) | simply means adding their respective -> 5 :(96+2553) î + (-7 -12013) il 7 components com Now. 151 = V(964255214+ (-7-12003) == 256.054) tano - Dyl -17 + 12013) 0 = tan post) tamo = =(96+2553 ) 10 = 57.0415

I have taken approximated values of sin(53) and cos(53). Please use calculator if you want the exact answer without approximation.

D = 256.054 Newton 57.0415 degrees below x axis.

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