Question: In the common setup shown in Figure 1, the hanging cylinder of mass m is released from rest. FALLING BLOCKS ON A STRING PHYSICS HOW TOSo the acceleration of the cart is = (mg) /(M+m) And the tension in the rope is (Mmg) /(M+m) How to solve Pulley Tension Problems – setup 2 Now again from equation 2 and 4 we get a complete expression of the tension T in the rope. W2 – T = ma mg – T = ma ……………….(3) Now combining equation 2 and 3, we get mg – Ma = ma or, a = (mg) /(M+m) -–(4) Here in equation 4, we get the expression of the acceleration of the cylinder and the cart. Only one equation and that is along Y-axis. cylinder – equations to solve pulley tension problems Therefore these two forces are not balanced and there is a net force acting on the cylinder which causes an acceleration of it downwards. – The tension force T also works on the cylinder through the string tied with it. This weight mg works vertically downwards. – Gravity or the weight W2 which is equal to mg. FALLING BLOCKS ON A STRING PHYSICS FREEPulley in Physics – Three forces on the cart Cart – free body diagramįigure 3- FBD for the cylinder Two forces on the cylinder (figure : 3) The cart has no acceleration in the y-direction. The acceleration of the cart along x must equal the acceleration of the cylinder, so we have used the same symbol a for both. There is only one equation for the cylinder. – We must write two equations for the cart because there are forces along both x and y directions. After drawing the FBDs we have to apply Newton’s second law in component form. This choice of coordinates means that the acceleration of both objects is along a positive x-axis (to the right for the cart downward for the cylinder). The tight rope ensures that the acceleration of both the cylinder and the cart has the same magnitude. Coordinate systems with X-axis and Y axis were used to make calculations easy. The cart accelerates to the right when the cylinder accelerates downward. The acceleration a of each subject is indicated. This is generally a common consideration for pulley tension problems. Coordinate systems and Common acceleration – Pulley in Physicsįor an ideal pulley, the tension is the same throughout the rope (therefore the same symbol T in both diagrams).
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