Block sliding down a wedge, all resting on a balance

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The net force on the sliding block is directed down the incline, with its magnitude determined by the component of gravitational force acting parallel to the wedge's surface. The force exerted by the block on the wedge includes both horizontal and vertical components, influenced by the block's weight and the angle of the incline. For the wedge to remain stationary, the horizontal component of the block's force must be balanced by the reaction force from the ground. When the block is accelerating, the reading of the balance reflects the normal force, which is affected by the block's weight and the incline's angle. Understanding these forces is crucial for analyzing the dynamics of the system.
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Hello! I am now working on an exercise which is related to mechanics. I calculated it wrongly. Attached is the answer. I have no idea how to do c(ii), I don’t understand how cos theta is applied in that question. Please explain it to me if you get it. thanks a lot!
Relevant Equations
F=ma
a=mg*sin theta
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IMG_0131.jpeg
 
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Consider the wedge block sliding down on the rough incline. Please answer the following three questions which are an elaboration of the hint in c(ii).
  1. What is the magnitude and direction of the net force on the sliding block? It cannot be zero because the block is accelerating.
  2. What is the force exerted by the block on the wedge? It must be a 2D vector with horizontal and vertical components.
  3. What must be true for the wedge to remain stationary?
Bonus question: What is the reading of the balance when the block is accelerating?
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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