Forces ­ An Application of Vectors Worksheet Question 1 A body of mass 10kg is attached to one end of a cord, the other end of which is fixed to a peg in a wall. The body is pulled aside by a horizontal force so that the cord is inclined at an angle of 30° to the vertical. Find the horizontal force and tension in the string. Question 2 A body of mass 4 kg is attached to one end of a string, the other end of which is fixed to a peg in the wall. The body is held in equilibrium, with the string making an angle of 30° to the vertical, by a force of P newtons which acts on the body in a direction perpendicular to the string. Find the value of P and the tension in the string. Question 3 A smooth ring of mass 2 kg is free to slide on a ring passing through it and tied to two points on the same horizontal level. In the equilibrium position, the angle between the directions of the string is 56°. Find the tension in the string. Question 4 A body of mass 800 grams is attached to one end A of a string AB, the other end being fixed to a wall; the length of AB is 780 mm. The body is pulled aside by a horizontal force and held in equilibrium at a distance AC=300 mm from the wall. Show that the angle q between the direction -5 of the horizontal force and the direction AB is given by cos q =
and that 13
12 sin q =
. Hence find the horizontal force and the tension in the string.
13
Forces An Applications of Vectors Worksheet Page 1 Solutions – Forces – An Application of Vectors Question 1 r = ( P cos 0° + T cos120° + 100 cos 270 ° ) ˆ i %
+ (T sin120° + 100 sin270 ° ) ˆ j Since the system of forces is in equ ilibrium , P cos 0° + T cos120° + 100 cos 270° = 0 and T sin120° + 100 sin270° = 0 \ P ­ T ´ 0.5 = 0 Þ P = 0.5 T 100 and T ´ 0.866 ­ 100 = 0 Þ T =
0.866 Hence T » 115.5N and P » 57.7 N
Question 2 r = ( P cos 30° + T cos120° + 40 cos 270 ° ) ˆ i %
+ ( P sin30° + T sin120° + 40 sin270 ° ) ˆ j Since the system of forces is in equilibri u m, P cos 30° + T cos120° + 40 cos 270° = 0 and P sin30° + T sin120° + 40 sin270° = 0 \0.866P ­ 0.5 T = 0 Þ T = 2 ´ 0.866 P 40 and 0.5P + 0.866T = 40 Þ P =
0. 5 + 2 ´ 0.866 2 (
Hence P » 20.0N an d T » 34.6 N Question 3
r = (T cos 62° + T cos118° + 20 cos 270 ° ) ˆ i %
+ (T sin62° + T sin118° + 20 sin270 ° ) ˆ j Since the sy stem of force s is in equilibrium , T cos 62° + T cos118° + 20 cos 270° = 0 and T sin62° + T sin118° + 20 sin270° = 0 20 \T =
Þ T » 11.3 N
( sin 62 ° + sin118 °)
Forces An Applications of Vectors Worksheet Page 2 )
Question 4 0.30
5 \ sin q =
0.78
13 ¢
and q » 22° 37 sin q =
å Horiz : P cos 0° + T cos112°37¢ + 8 cos 270 °
å Vert : P sin 0° +T sin112°37¢ + 8 sin270 °
Since the system of forces is in equilibrium, then å H = åV
= 0 P ­ T sin22° 37¢ = 0 and T cos 22° 37¢ ­ 8 = 0 8 \T =
Þ T » 8.7 N cos 22° 37 ¢
\ P = T sin22° 37¢ Þ P » 3.3 N
Forces An Applications of Vectors Worksheet Page 3