Find Magnitude & Direction of Force P for Friction Question

[snshusat161]

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 \mu = 0.2. Find the magnitude and direction of the least force P at which the motion of the block will impend.

Homework Equations

frictional force = \mu. 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 \theta. Somebody help me to find that angle.

Answer: P = 161.7 N and \theta = 11.31 degree

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

 

[rock.freak667]

Well what two equations did you get?

 

[snshusat161]

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 =\mu N + 150 sin 60
= 145 N

For body B

R + P sin\theta = 100 ......(i)

\mu R + 145 = P cos\theta

= 0.2 R - P cos\theta = -145 ...(ii)

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

 

[snshusat161]

bump!

 

[tiny-tim]

hi snshusat161!

(have a mu: µ and a theta: θ )

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

 

[snshusat161]

hi snshusat161!

(have a mu: µ and a theta: θ )

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

then I will have 1.01P Cos (\theta - 78.96) = 165

and still i have two unknowns and only one equation.

 

[snshusat161]

that 60 degree is not the value of \theta. It is unclear in diagram but \theta is the angle made by force P on the horizontal. (block A).

 

[snshusat161]

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 =\mu N + 150 sin 60
= 145 N

For body B

R + P sin\theta = 100 ......(i)

\mu R + 145 = P cos\theta

= 0.2 R - P cos\theta = -145 ...(ii)

Now I can solve this two equations and can calculate the value of P if I know value of \theta. 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 =\mu N + 150 sin 60
= 145 N

For body A

R + P sin\theta = 100 ......(i)

\mu R + 145 = P cos\theta

= 0.2 R - P cos\theta = -145 ...(ii)

 

[tiny-tim]

(just got up :zzz: …)

(what happened to that θ i gave you? )

then I will have 1.01P Cos (\theta - 78.96) = 165

and still i have two unknowns and only one equation.

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? ​

 

[snshusat161]

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.
i am on phone so I'm inclined to use texting language. please ignore if you feel that my posts are breaking any rules.