Bộ tài liệu bài tập tham khảo môn vật lý bậc đại học bằng tiếng anh Chapter 05 | a The tension in the rope must be equal to each suspended weight N. b If the mass of the light pulley may be neglected the net force on the pulley is the vector sum of the tension in the chain and the tensions in the two parts of the rope for the pulley to be in equilibrium the tension in the chain is twice the tension in the rope or N. In all cases each string is supporting a weight w against gravity and the tension in each string is w. Two forces act on each mass w down and T w up. a The two sides of the rope each exert a force with vertical component T sin Q and the sum of these components is the hero s weight. Solving for the tension T T I91 1 kgH9 S0m s I n . 2 sin 0 2 sin b When the tension is at its maximum value solving the above equation for the angle Q gives e arcsinf I arcsinf s2 i . 2TJ I 2 x 104 N J The vertical component of the force due to the tension in each wire must be half of the weight and this in turn is the tension multiplied by the cosine of the angle each wire makes with the vertical so if the weight is w w cos Q and Q arccosf 4S . With the positive y-direction up and the positive x-direction to the right the free-body diagram of Fig. b will have the forces labeled n and Tresolved into x- andy-components and setting the net force equal to zero Fx T cos a - n sina 0 Fy ncosa Tsina - w 0. Solving the first for n T cot a and substituting into the second gives cos a . cos2 a sin2 a T T---------f T sin a T I----1-------I ----- w sina sina sina J sina and so n T cot a w sin a cot a w cos a as in Example . w sin a mg sin a 1390 kg m s2 sin x 103 N. x rp TXT rT1 TXT 1 4090kg s r rs -s -1 z- 4 at .7 a Tb cos 0 - W or TB - W cos 0 -- 40 5-23 x 10 N. b Ta - Tb sin0 - x 104N sin 40 - x 104 N. a TC - w TA sin 30 Tb sin45 - TC - w and Ta cos30 - Tb cos45 - 0. Since sin 45 - cos45 adding the last two equations gives Ta cos30 sin30 - w and so T - - .
