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`        Determine the horizontal force P to be applied to a block of weight of 1800N to hold it in position on a smooth inclined plane, which makes an angle 300 with horizontal reference line.  `
3 months ago

Vikas TU
9517 Points
```							Dear student Weight downward = 1800N Horizontal force required to keep Block in inclined plane = mgSin30 mg = 1800N Horizontal force = 1800*1/2 => 900 N Hope this helps
```
3 months ago
Joel Sudhakar
23 Points
```							dear brother here weight acting downward =1800Nthe force acting  along the plane to hold the block is =mgSin30=1800*1/2=900but here we need to calculate the horizontal force ‘P’which is acting on the body so, we resolve the horizontal force ‘P’ into x-component and y-component as PCos30 and Psin30 respectivelythen Pcos30=mgSin30P=(1800*1/2)/Cos30=1039.23N
```
3 months 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