Vector Addition: Find the Resultant Vector

  • #1
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Homework Statement


[/B]
The block shown in Fig. P-223 is acted upon by its weight W = 200lb, a horizontal force Q = 600lb and the pressure P exerted by the inclined plane. The resultant R of these forces is up and parallel to the incline thereby sliding the block up it. Determine P and R. Hint: Take one axis parallel to the incline. Ans: R = 293lb

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t1.0-9/q82/s720x720/11015301_1002403293112217_141332182178751666_n.jpg?oh=13938518c6c5618236be3f0fbd8ce4aa&oe=562F0B00&__gda__=1445053718_99ad9a3e011c3cbad300e49a737ea336





Homework Equations


[/B]R = SummationFx + SummationFy
Fx = Fcos0
Fy = Fsin0


The Attempt at a Solution


I don't know how to start the problem. In the worktext, this is the only problem that has unknown magnitude(P). To get the R I need first to P. The hint says take one axis parallel to the incline, I will be able to form an isosceles triangle with P as the leg but I still won't be able to solve for the magnitude of P.... How do I start this problem?


This problem is from Engineering Mechanics by Singer, it is our homework #213-223, I'm already done with the rest. The previous problems has all given magnitude and angle(some you need to solve for the angle, but still are easy except this one because of unknown value of P).
 
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Answers and Replies

  • #2
Nathanael
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Uhh, there's no way to find the pressure P without knowing the area of the bottom of the box. Perhaps they meant to say that P is a force.

With that assumption, that P is a force, you can determine it's magnitude by the fact that the block is in equilibrium in the normal direction (it doesn't float away from the plane and it doesn't move inside the plane).
 
  • #3
SammyS
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Homework Statement


[/B]
The block shown in Fig. P-223 is acted upon by its weight W = 200lb, a horizontal force Q = 600lb and the pressure P exerted by the inclined plane. The resultant R of these forces is up and parallel to the incline thereby sliding the block up it. Determine P and R. Hint: Take one axis parallel to the incline. Ans: R = 293lb

https://fbcdn-sphotos-h-a.akamaihd.net/hphotos-ak-xpf1/v/t1.0-9/q82/s720x720/11015301_1002403293112217_141332182178751666_n.jpg?oh=13938518c6c5618236be3f0fbd8ce4aa&oe=562F0B00&__gda__=1445053718_99ad9a3e011c3cbad300e49a737ea336

Homework Equations


[/B]R = SummationFx + SummationFy
Fx = Fcos0
Fy = Fsin0

The Attempt at a Solution


I don't know how to start the problem. In the worktext, this is the only problem that has unknown magnitude(P). To get the R I need first to P. The hint says take one axis parallel to the incline, I will be able to form an isosceles triangle with P as the leg but I still won't be able to solve for the magnitude of P.... How do I start this problem?

This problem is from Engineering Mathematics by Singer, it is our homework #213-223, I'm already done with the rest. The previous problems has all given magnitude and angle(some you need to solve for the angle, but still are easy except this one because of unknown value of P).
The vector P may be unknown, although you do know its direction.

However, you do know something about the resultant R. If you follow the hint, then you know on of the components of R. Right ?
 
  • #4
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@sammy:

Great!!! I got it now thanks. What I did is I made the incline my x-axis (hint). With that I now have:

Q = 600∠-30
W = 200∠-120

and then:

Qx = 519.62
Qy = -300
Wx = -100
Wy = -173.21

and since R is parallel to the incline which is now the x-axis therefore ΣRy = 0 = Qy + Wy + Py or

Py = 473.21... using the given 15' angle, I get Px = -126.8

R = ΣRy + ΣRx
R = ΣRx
R = Qx + Wx + Px

R = 292.82


Sweet!!! Thanks a lot :)
 
  • #5
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Can you show the solution on how did you get the value of Py and Px?
 
  • #6
haruspex
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Can you show the solution on how did you get the value of Py and Px?
You should post your own attempt first.
 

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