Solving the system of equations using Cramer's rules, we obtain The above is a system of 2 equations with 2 unknown |T 1| and |T 2|. T 1 + T 2 + T 3 = 0 (Newton's second law) T ' 3 = (0, |T ' 3|), tension of stringĪt point p (origin of x y system of axes) Problem 4 Three cords are knotted at point P, with two of these cords fastened to the ceiling making angles α1, α2 and a block of mass m hangs from the third one as shown below.Ī) Find the magnitude of the tension in each cord in terms of α1, α2 and m so that the system is at rest.ī) Find numerical values to the three tensions found above for α1 = 45°, α2 = 30° and m = 1 Kg. Substitute in the above equation to write W 1 and W 2 are the weights of m 1 and m 2 and are given by Using the fact that |T 1| = |T 2| and the last two equations to write T 1 and T 2 represent the tension in the same string and therefore their magnitudes are equal. Use equation (1) and (2) found above to write W 2 + T 2 = m 2 a 2 ( Newton's second law (vector equation))Ī 2 = (0, - |a|), assuming m 2 accelerating downward. T 2 the tension of the string at m 2 (force exerted by string on m 2) W 1 + T 1 = m 1 a 1 ( Newton's second law (vector equation) )Ī 1 = (0, |a|) acceleration assuming m1 accelerating upward. T 1 the tension of the string at m 1 (force exerted by string on m 1) Let |a| the magnitude of the acceleration of m 1 and m 2 assuming m 1 accelerating upward and m 2 accelerating downward. Find an expression of the acceleration when the block are released from rest. In the two blocks of masses m 1 and m 2 and pulley system below, the pulley is frictionless and massless and the string around the pulley is massless. Use |T 1| = m 1 |a| and |a| = m 2 g / (m1 + m 2) to obtain Note: |T 1| = |T 2| tension in the string is the sameĬombining the equations |T 1| = m 1 |a|, |T 1| = |T 2| and -m 2 g + |T 2| = - m 2 |a| found above, we can write Newton's second law, assuming m 2 accelerating downward and |a| is the magnitude of the acceleration Newton's second law, assuming m 1 accelerating from left to right and |a| is the magnitude of the acceleration.Ībove equation in components form: (0, -m 1 g) + (0, |N|) + (|T 1|, 0) = (m 1 |a|, 0) We assume that the string is massless and the pulley is massless and frictionless.Ī) Find the magnitude of the acceleration of the two masses Problem 2 In the figure below is shown the system below are shown two blocks linked by a string through a pulley, where the block of mass m 1 slides on the frictionless table. T 2 and F c are action and reaction pairs and therefore their magnitudes are equal. T 1 and T 2 represent the tension of the string and their magnitudes are equal. Sum of y coordinates = 0 gives |W| = |T 1| Since the block is at rest W + T 1 = 0 ( Newton's second law, vector equation) We now consider the forces acting on the block (Free Body Diagram) Hence action reaction (Newton's 3 rd law) : |F c| = |T 2| Tension T 2 acting on the ceiling and F c the reaction to T 2. Assume the mass of the string to be negligible.įree body diagram below shows the weight W and the tension T 1 acting on the block. Find the force F c exerted by the ceiling on the string. Problem 1 A block of mass 5 Kg is suspended by a string to a ceiling and is at rest. Problems involving forces of friction and tension of strings and ropes are also included. Free body diagrams of forces, forces expressed by their components and Newton's laws are used to solve these problems. Several problems with solutions and detailed explanations on systems with strings, pulleys and inclined planes are presented.
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