How To Solve Tension Problems

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Along Y axis there is no net force, so N – W1 =0 or, N – Mg =0or, N = Mg …………………..(1)Along X axis, the tension force T gives an accelerated motion to the cart, T = Ma …………………….(2) – Gravity or its weight W2 which is equal to mg. If the friction between the cart of mass M and the horizontal track is present with coefficient of friction μ , find the acceleration of the cart and the tension in the rope.

In the previous set up, there was no friction between the cart’s wheels and the table surface below.

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Now suppose I want to find the acceleration of the cart after it is let go.

First, the string makes the magnitude of the acceleration for both carts is the same.

This is basically just series of lines on the track that the cart uses to determine its position (and thus velocity and acceleration).

With a cart mass of 1207 grams (mass 1) and a hanger mass of 145 grams (mass 2), I get the following data. Why is the tension not the same as the weight of mass 2?

is one of the most interesting topics in mechanics. 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).

Once you understand the application of Newton’s laws in pulley systems, it may become one of the most favorite topics for you. After drawing the FBDs we have to apply – We must write two equations for the cart because there are forces along both x and y directions. The acceleration of the cart along x must equal the acceleration of the cylinder, so we have used the same symbol a for both. – gravity or weight of the cart W1 = Mg, working vertically downwards– the normal force N, vertically upwards– and the tension force T acting through the rope along positive x axis, as shown above. Therefore these two forces are not balanced and there is a acting on the cylinder which causes an acceleration of it downwards. W2 – T = mamg – T = ma ………………..(3)Now combining equation 2 and 3, we getmg – Ma = maor, a = (mg) /(M m) ——————–(4) Here in equation 4, we get the expression of the acceleration of the cylinder and the cart.


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