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Force needed to pull a block up an incline. |
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| Sep30-09, 03:13 PM | #1 |
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Force needed to pull a block up an incline.
1. The problem statement, all variables and given/known data
Calculate the force needed to pull a mass of 20 kg at a uniform slow speed up a plane inclined at an angle of 30 with the horizontal if the coefficient of kinetic friction is 0.20. 2. Relevant equations WN= w cos [tex]\vartheta[/tex] WT= w sin [tex]\vartheta[/tex] [tex]\mu[/tex]s= tan[tex]\vartheta[/tex] 3. The attempt at a solution I don't even know how to get started. |
| Sep30-09, 07:38 PM | #2 |
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If the block is moving at a constant speed then you know that there is no net force acting on the block.
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| Sep30-09, 08:56 PM | #3 |
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so the friction force is
(.2)(20)(9.8)(cos 30) ? and the gravitational force is (20)(9.8)(sin 30) ? i got 34 N for friction and 98 N for gravitational. are they supposed to equal zero? or do i add them together to find the force i need to overcome? or could i just overcome the strongest? |
| Oct1-09, 05:31 AM | #4 |
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Force needed to pull a block up an incline.
So what force is needed to make them balance?
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| Oct1-09, 08:26 AM | #5 |
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okay, so 34 N for friction pull the box up the slope, and 98 N gravity pull down.
98 N - 34 N = 64 N needed to equalize them, and more than 64 to make it move uphill? |
| Oct2-09, 07:22 AM | #6 |
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What direction will the friction be acting in if the box is being pulled up the slope?
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| Oct2-09, 07:49 AM | #7 |
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Recognitions:
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What is the direction of the force applied and whether the friction will depend on that direction.
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| Oct2-09, 07:57 AM | #8 |
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Recognitions:
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If the minimum force needed is required than the force must be applied at an angle equal to angle of friction [tan^-1 (u)] with the incline.
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