Finding acceleration of a block being pushed up a ramp by a horizontal force

  • #1

Homework Statement


A block is pushed up a frictionless incline by
an applied horizontal force as shown.
The acceleration of gravity is 9.8 m/s2 .
What is the magnitude of the resulting acceleration of the block?
Answer in units of m/s^2.


Homework Equations


Sin(theta)=opposite/hypotenuse
Cos(theta)=adjacent/hypotenuse
Tan(theta)=opposite/adjacent
^^^^^^^^^above equations for a right triangle on a Free Body Diagram
F=ma

The Attempt at a Solution


So if i extend the block's line and make a right triangle with that line and the line of the applied force, i know the theta of that triangle is equal to the ramp's, 34. I can also make a right triangle using gravity and its two vectors, one along the ramp and one perpendicular to the ramp. That triangle's theta is also 34. So to find the acceleration of the block, i need to subtract x from v (See attached picture). But i can't prove the two triangles are congruent, because the gravity vector triangle has only two known values while the force vector triangle has only 1. Help!
 

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Answers and Replies

  • #2
Wait, i got a little further on the problem.
Knowing that gravity is 9.8, i can find x and y of the gravity vector triangle.
x=sin(theta)g=~5.480090454
y=cos(theta)g=~8.124568211
SO now i know x.
I need to find v to get the net force, how?
 
  • #3
Doc Al
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I need to find v to get the net force, how?
The same principles that apply to one right triangle will apply to another. If you can relate mg to y, you can relate z to v.
 
  • #4
ok i used the formula to calculate v and x. Then using f=ma, i found the acceleration. I was wrong.
I found a mistake, in that in the gravity vector triangle, i used 9.8 which is in m/s^2. After converting it to newtons, i get that x is 20 something newtons, which cannot be correct, because then the net force is negative, meaning the block is going down the ramp. However, the problem states the block is going up the ramp. whaaaaat?
 
  • #5
Doc Al
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Yes, given the values in the diagram, the net force is down the plane. Are you sure the problem stated that the block is moving up the plane? Or did it just say that it was being pushed up the plane (meaning that the push has a component up the plane)?
 
  • #6
Well, i took x (11.18385807)-v (30.06090238) and got -18.87704431 N. Then i divided by 3.7 (the mass) and got 5.101903868, which is wrong. What?
 
  • #7
Doc Al
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Well, i took x (11.18385807)-v (30.06090238) and got -18.87704431 N.
Your values for x and v are incorrect.
 
  • #8
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0
i agree that the answer is a net force acting down the slope of the hll
 
  • #9
are the correct values
13.746065934159120354149637745012 for x
and
5.4800904540133189355721957718627 for v?
 
  • #10
Doc Al
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