Inclined Plane phyiscs question

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Homework Statement


He gives the bike a little throttle providing an applied force of 1.20 kN to propel him up the 25 degree incline. THe combined weight of Chad and the dirt bike is 1.85 kN. If the force of friction is 250 N, determine the acceleration?


2. Homework Equations [/b
I used Fnetx(ma)= Ff + Fa + Fg ( parallel)


The Attempt at a Solution


I found the mass by doing Fg / g to get mass which was 188.58 kg. so I know the Ff + Fa + Fg(parrallel) will be divided by this mass. I know Ff is -250. the applied force is 1200N. So what is the Fg( parallel). Or did i even do this right??
 

Answers and Replies

  • #2
gneill
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I think you need to draw a free body diagram for the situation, and examine how the geometry determines the downslope component of the weight.
 
  • #3
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I did draw the free body diagram Fg is straight down Fg perpendicualr is straight down from the surface and Fn is straight up from the surface, and Fg having a horizontal component called Fg (parallel) and Ff going left from the bike on the inclined plane and the applied force is opposite.
Gneill, I don't understand what you are saying, why cant anyone just sohw me how to get the answer and what im getting wrong?
 
  • #4
gneill
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It would be nice to see some calculations with values. You've determined the direction of the frictional force, which is good. You've also got the mass from the weight. It seems that what you're missing is the downslope component of the weight. Your FBD should allow you to find it from the geometry of the situation.
 
  • #5
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the calcualtions only take one step. -250+1200sin25+1200N divided by 188.58 Kg to get acceleration, but I got it wrong on the test
 
  • #6
gneill
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the calcualtions only take one step. -250+1200sin25+1200N divided by 188.58 Kg to get acceleration, but I got it wrong on the test

I take it that -250 is the downslope frictional force, and I can spot the upslope force of the bike propulsion, 1200N. What is 1200sin25 supposed to represent? It's positive so it's an upslope-directed force. Doesn't make sense to me.
 
  • #7
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it represents the Fg(parallel) I remember when my teacher was going over this question he put a number then sin25, i just cant remember the number.
Never mind, this question is hopeless
 
  • #8
rl.bhat
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Applied force in in the upward direction. The frictional force and proponent of the weight of the Chad and the bike (1850*sin25) are in the downward direction. Find the net upward force and hence acceleration.
 
  • #9
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i don't understand, you dont need Fn. Because the Fnet equation equals Ff + Fa + Fg(parallel) , you dont need Fn or Fg becuase its in the y-axis, and acceleration is in the x-axis
 
  • #10
rl.bhat
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I hope the applied force on the bike is parallel to the inclined plane. The bike experiences a retarding force ( fr + W*sin25) where W is the weight of the bike and Chad. Find the net force and proceed.
 

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