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The diagonal arrows are forces from normal reaction. I didn't wrote it. Sorry for any inconveniences.sahilmm15 said:I have drawn all the Free body diagrams finding difficulty in equating them
Without some stated notation, I'm not quite sure what your diagrams mean. You show some actual forces, like mg, but also resultants, like ma, and omit to label the normal forces. It would be clearer had you got as far as writing equations.sahilmm15 said:The diagonal arrows are forces from normal reaction. I didn't wrote it. Sorry for any inconveniences.
Well , after some time (30 minutes) I was able to solve the problem. I completely left the problem for 1 hour so that my brain can work on it subconsciously(as suggested by one of my mentors in other thread). It came to my surprise that I solved it easily. Shall I provide you my work?kuruman said:Drawing free body diagrams (FBDs) doesn't get you very far unless you write equations based on Newton's second law appropriate to them. You have correctly identified that you have three systems, the hanging mass, the wedge and the block. What equations can you get out out of them? Hint: ##\vec F_{\text{net}}=m\vec a.## Write expressions for each of the left and right sides for each of the 3 systems.
VEReade said:1)You're free body force diagram (FBD) for falling mass M looks good.
2)The other two don't look right. For example, the triangular mass M' is accelerating with same acceleration as M to the right. There is also a reaction force on M' due to mass m to consider.
3)To solve this problem, you need to think about what it means acceleration-wise for the m not to slipping over M'
4)Note, in these kind of questions, although you'll always be applying 'F=ma", there will often be more than one way of getting the right answer!
Also for triangular block I didn't draw the fbd but directly wrote the equation.sahilmm15 said:In step 2 of my answer it states "to keep 'm' at rest".And the statement I wrote in step 1 was for step 2. Sorry for any inconveniences.
The final step is wrong.sahilmm15 said:##M=M\tan(\theta)+(M'+m)\tan(\theta)##
##M=\frac{M'+m}{\cot(\theta)}##
Thanks for your reply, I appreciate it. It should be cot(x) - ##1## in the denominator. It was an calculation error.haruspex said:The final step is wrong.
Yes.sahilmm15 said:Thanks for your reply, I appreciate it. It should be cot(x) - ##1## in the denominator. It was an calculation error.
A classic string and pulley problem is a physics problem that involves a system of one or more pulleys and a string or rope. The goal of the problem is usually to determine the forces acting on the pulleys and the tension in the string.
To solve a classic string and pulley problem, you will need to use the principles of mechanics and Newton's laws of motion. You will also need to draw a free-body diagram and apply equations of motion to the system of pulleys and strings.
Some common assumptions made in a classic string and pulley problem include assuming the pulleys are massless and frictionless, the string is inextensible, and the system is in equilibrium.
Yes, there are different methods that can be used to solve a classic string and pulley problem, such as using vector analysis, trigonometry, or algebraic equations. The method used will depend on the specific problem and the individual's preference.
Classic string and pulley problems have many real-life applications, such as in elevators, cranes, and other lifting mechanisms. They are also used in physics experiments to study the principles of mechanics and in engineering to design efficient and effective systems.