Christmas MT's answer to Angelica's Junior College 1 H2 Maths question.
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In part (a), you have equilibrium. That means net horizontal force and net vertical force at Q are both 0.
In part (b), all the horizontal force acting on the left side is removed by the thread being cut. Since the force in the R direction does not change, the net vertical force remains 0. However, the net horizontal force is now equal to the horizontal component of the force in the direction of R
In part (b), all the horizontal force acting on the left side is removed by the thread being cut. Since the force in the R direction does not change, the net vertical force remains 0. However, the net horizontal force is now equal to the horizontal component of the force in the direction of R
Date Posted:
7 years ago
Thanks alot but can I just ask why we do t have to consider the tension acting in the string holding on to object S?
We do consider tension in the string holding on to the object. The tension is equal to the weight of the object
Let's revisit the forces at O. There is tension from three strings (the tension from the string holding on to the object is equal to the weight of the object). The sum of the vertical and horizontal components of the tensions is equal to 0.
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