Question

+ to page Pageview Adidated 4. The frame shown is subjected to a horizontal force F=300j N. (a) Find the vector component of F parallel to the bar AB (3) A F-MON 3 m (b) Obtain a vector perpendicular to both F andra (the position vector from A to B). (3) 5. The crate is held in place on the smooth inclined surface by the rope AB. The tension in AB is 70 N. Determine the mass of the box and the normal force exerted on it by the inclined surface. A free bowly diagram is essential. (5) BU Type here to sewich

Q4 please asap

Answer 1

Kindly go through the solution provided below.

1 8002 i + 6î +3K) We are given that F = 300j (2,6,3) BF-70 N AB (0,0,0) a) 36 = (2-0)i + (6-0); + (3-0) ☆ Ov 316 = 2+ 69 +31 Now, the vector component of F farallel to tas is given by the dot product of Ŝ and the 21 + 6 + 3 21 +6j + 3 ľ +6²+ 3² uit vector î Fas ㅋ 1800 7 - F ñ = (360 j). (27 + 6ý +36) ㅋ que so; ㅋ ㅋ 18.00 (Ri + 6; +3*) 49 b. Also, a vector perpendicular to both Ĉ and % en by the cross froduct F and PAB is given of k - 300 Ex Bab= 0 o 3 6 2 1990-0)-1 (0-0) + (0 - 6 lll lll đo 3 - 6 Gok