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`        A block of mass 0.5 kg rests on the 8nclined surface of a wedge of 2 kg.  The wedge is acted on by a horizontal force F and slides on a frictionless surface . If the coefficient of static friction between the wedge and the block is 0 8 and the angle of inclination is 37° , maximum value of F for which the block doesn't slip is approximately ________ ?`
one year ago

## Answers : (1)

Arun
23345 Points
```							Fn - mgcos = 0" Not this time. The normal rules don't apply. The whole system is accelerating. The direction of the y axis you're using in that equation isn't "up". Find the component of a that's in the direction of your y axis. That's a sin theta. Fn - mg cos theta = ma sin theta "-mgsin + ff = m2a (x)" I guess you realized something was fishy at this point, since you wrote "(x)". Problem is, after you wrote it, you ignored it. Same deal as above. Find the component of a that's in the direction of your x axis. -mg sin theta + ff = -ma cos theta Note the minus sign on the right because I'm assuming acceleration is opposite the direction of your x axis. It doesn't matter if I'm wrong - you will get a negative result for a, and just switch it. Solve for a, then you can easily using eqn. 1 to find the force. Also, now you need to find the maximum and minimum values for F, so also write -mg sin theta - ff = -ma cos theta And solve again. In this case the friction is actually keeping the block from sliding up the wedge, like dishes on a table when you pull the tablecloth out. I think the kg is just a mistake in the book. The correct answer should be
```
one year ago
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### Course Features

• 110 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions