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Question

This question is from an online homework website, where we getassigned homework weekly. The question is:
The coefficient of static friction between the 3.68 kg crateand the 22°incline is 0.277. The acceleration of gravity is 9.8 m/s^2. Whatminimum force F must be applied to the crate perpendicular to theincline to prevent the crate fro sliding down the incline? Answerin units of N.
Then, there is a picture,which isn't exactly needed as there is no extra info in thepicture. But, it is a picture of an incline with the 22°labeled. There is a box on top of the incline labeled 3.68 and anarrow pointing perpendicular to the box labeledF.
The coefficient of static friction between the 3.68 kg crateand the 22°incline is 0.277. The acceleration of gravity is 9.8 m/s^2. Whatminimum force F must be applied to the crate perpendicular to theincline to prevent the crate fro sliding down the incline? Answerin units of N.
Then, there is a picture,which isn't exactly needed as there is no extra info in thepicture. But, it is a picture of an incline with the 22°labeled. There is a box on top of the incline labeled 3.68 and anarrow pointing perpendicular to the box labeledF.

Explanation / Answer

In this physical situation, the perpendicular force, F toprevent the weight of the block from sliding, will be the staticfrictional force, fs : Now fs =ukmgcos We compute using Newtons 2ndlaw:        Since we are on anincline, the weight, mg has a force equal mg sin, which isthe force that would start the block slidingr; now there is noacelleration , thus we write:                               mgsin - fs = m a subbing:                                     FB -f s = 0            Now,                                     fs = (0.277)(3.68)(9.8)(0.927)                                       = 9.26 N This perpendicular force equal the frictional staticforce. The force of the block weight is: FB =(3.69)(9.8)(0.375) = 13.55 N This is greater than the frictiona force, so the block wouldstart sliding unless more perpendicular force is applied . To find this we take the difference:    13.55 N- 9.26 N = 4.29 N          So 4.29N must be applied perpendicular to the block to prevent it fromsliding.                                                                                                          

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