Finding minimum force P and angle for impending motion with connected blocks

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Homework Statement



Two rectangular blocks of weight, A = 150 N and B = 100N are connected by a string and rest on an inclined plane and on a horizontal surface as shown in the figure. The coefficient of friction for all contiguous surfaces is [tex]\mu[/tex] = 0.2. Find the magnitude and direction of the least force P at which the motion of the block will impend.
attachment.php?attachmentid=29400&stc=1&d=1288005670.jpg



Homework Equations




frictional force = [tex]\mu[/tex]. N

where N = normal rection


The Attempt at a Solution



I had calculated T In the string from box A and then I have made two equations from box B. The only think I am unable to find is the angle [tex]\theta[/tex]. Somebody help me to find that angle.

Answer: P = 161.7 N and [tex]\theta[/tex] = 11.31 degree

sorry for the poor diagram. I don't have a scanner.
 

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Hard to show me in the diagram but I know you all can understand cause you all are very well known to such resolution of forces. So I'm proceeding directly without any diagram.

For body A

N = 150 cos 60 = 75 N

T =[tex]\mu[/tex] N + 150 sin 60
= 145 N

For body B

R + P sin[tex]\theta[/tex] = 100 ......(i)

[tex]\mu[/tex] R + 145 = P cos[tex]\theta[/tex]

= 0.2 R - P cos[tex]\theta[/tex] = -145 ...(ii)

Now I can solve this two equations and can calculate the value of P if I know value of [tex]\theta[/tex]. I want your help.
 
hi snshusat161! :smile:

(have a mu: µ and a theta: θ :wink:)

subtitute for R from (i) into (ii), and your equation should be of the form P(Acosθ + Bsinθ) = C …

now write Acosθ + Bsinθ as a multiple of cos(θ+φ), for some φ :smile:
 
tiny-tim said:
hi snshusat161! :smile:

(have a mu: µ and a theta: θ :wink:)

subtitute for R from (i) into (ii), and your equation should be of the form P(Acosθ + Bsinθ) = C …

now write Acosθ + Bsinθ as a multiple of cos(θ+φ), for some φ :smile:


then I will have 1.01P Cos ([tex]\theta[/tex] - 78.96) = 165

and still i have two unknowns and only one equation. :confused:
 
that 60 degree is not the value of [tex]\theta[/tex]. It is unclear in diagram but [tex]\theta[/tex] is the angle made by force P on the horizontal. (block A).
 
snshusat161 said:
Hard to show me in the diagram but I know you all can understand cause you all are very well known to such resolution of forces. So I'm proceeding directly without any diagram.

For body A

N = 150 cos 60 = 75 N

T =[tex]\mu[/tex] N + 150 sin 60
= 145 N

For body B

R + P sin[tex]\theta[/tex] = 100 ......(i)

[tex]\mu[/tex] R + 145 = P cos[tex]\theta[/tex]

= 0.2 R - P cos[tex]\theta[/tex] = -145 ...(ii)

Now I can solve this two equations and can calculate the value of P if I know value of [tex]\theta[/tex]. I want your help.


Sorry, I made a mistake here.

Here's the correct one

For body B

N = 150 cos 60 = 75 N

T =[tex]\mu[/tex] N + 150 sin 60
= 145 N

For body A

R + P sin[tex]\theta[/tex] = 100 ......(i)

[tex]\mu[/tex] R + 145 = P cos[tex]\theta[/tex]

= 0.2 R - P cos[tex]\theta[/tex] = -145 ...(ii)
 
(just got up :zzz: …)

(what happened to that θ i gave you? :confused:)
snshusat161 said:
then I will have 1.01P Cos ([tex]\theta[/tex] - 78.96) = 165

and still i have two unknowns and only one equation. :confused:

ok, if that was the correct equation, what it would tell you?

it says that you can get the desired motion for different values of θ, and for each value of θ there's only one value of P …

but for what value of θ is that value of P a minimum? :smile:
 
sorry it was sine not cosine. so p will be minimum for max value of sine. i.e theta minus phi is equal to ninety. thanks tiny tim you really explained me in a very interesting nice and smart way.
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