Force and Motion with two unknows

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The discussion revolves around a physics problem involving two bodies on an incline, focusing on the forces acting on them, including tension and friction. The participants are exploring the acceleration of body A under different conditions while considering the forces acting on both bodies.

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  • Mixed

Approaches and Questions Raised

  • Participants discuss the equations of motion for both bodies, noting the presence of two unknowns: tension and acceleration. Some express difficulty in solving the problem and seek clarification on the equations presented.

Discussion Status

There is ongoing dialogue with participants sharing their equations and attempting to clarify their reasoning. Some have provided equations for both bodies, while others are struggling to follow the logic and are asking for further elaboration.

Contextual Notes

Participants are working under the constraints of the problem's parameters, including weights, coefficients of friction, and angles, while grappling with the challenge of multiple unknowns in their equations.

ankitp3
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Body A in Fig. 6-36 weighs 100 N, and body B weighs 77 N. The coefficients of friction between A and the incline are μs = 0.52 and μk = 0.28. Angle θ is 31°. Let the positive direction of an x-axis be down the slope. What is the acceleration of A if A is initially (a) at rest, (b) moving up the incline, and (c) moving down the incline?
http://edugen.wiley.com/edugen/courses/crs1650/art/qb/qu/c06/fig06_34.gif

=S m stuck @ -FT + mgsintheta = ma and i am stuck with two unknowns.. Tension and acceleration
 
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Welcome to PF!

ankitp3 said:
-FT + mgsintheta = ma and i am stuck with two unknowns.. Tension and acceleration

Hi ankitp3! Welcome to PF! :smile:

Yes, you do have two unknowns.. tension and acceleration …

but you also have two equations, one for body A and one for body B. :wink:
 


still having hard time solving this problem :S
 
ankitp3 said:
still having hard time solving this problem :S

ok, show us your two equations, and we'll see what we can do … :smile:
 
FK + mgsingtheta - FT = m1ax (block A)

Ft - Fg = m2gy (block b)

ukmgcostheta - mgsintheta - FT = m1ax

Ft - (wB) = (wB/9.8)a

ukmgcostheta + mgsintheta - Ft (< sub in Ft from the last equation)
 
ankitp3 said:
FK + mgsingtheta - FT = m1ax (block A)

Ft - Fg = m2gy (block b)

ukmgcostheta - mgsintheta - FT = m1ax

Ft - (wB) = (wB/9.8)a

ukmgcostheta + mgsintheta - Ft (< sub in Ft from the last equation)

Sorry … I've stared at this for several minutes and I can't follow it. :confused:

Can you put the numbers in? :smile:
 

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