How to Solve a Friction Problem with Statics

[Jud]

Homework Statement

Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 35kg and the coefficient of static friction between the crate and the plane is μs = 0.40

2. The attempt at a solution

Fx = P cos 30 - 343.35 sin 30 + F = 0
Fy = -P sin 30 - 343.35 cos 30 + N = 0

F = (μs)(lNl) = 0.40 (N)

so,

Fx = P cos 30 - 343.35 sin 30 + 0.40(N)

therefore,

Fy = -P sin 30 - 343.35 cos 30 = -N

so, F = 0.40 (P sin 30 + 343.35 cos 30)Sub into Fx,

P cos 30 - 343.35 sin 30 + 0.40 (P sin 30 + 343.35 cos 30) = 0Now solving for P I get 51.6 N, which is incorrect.
The answer is given. P = 49.5 N

Can anyone give me an indication of where I have gone wrong please.

Thank You.

 

[voko]

You went wrong when you plugged numbers in your equations very early. Work out the result symbolically, then plug in the numbers.

 

[Jud]

P cos θ - mg sin θ + μs ( P sin θ + mg cos θ) = 0

Still yields the same answer?

 

[voko]

This is not the result, this is just a step toward it. The result would have the form P = ... where the right hand side would not contain P.

 

[Jud]

P ( cos θ - mg sin θ + μs (sin θ + mg cos θ)) = 0

I'm having major trouble transposing.

P(-mg sin θ + μs (sin θ + mg cos θ)) = -cos θ

P(μs(sin θ + mg cos θ)) = -cos θ + mg sin θ

P(sin θ + mg cos θ) = -cos θ + mg sin θ / μs

P (sin θ) = (-cos θ + mg sin θ / μs) - mg cos θ

P = (-cos θ + mg sin θ / μs sin θ) - mg cos θ)

 

[voko]

P cos θ - mg sin θ + μs ( P sin θ + mg cos θ) = 0

Open the brackets.

P cos θ - mg sin θ + μs P sin θ + μs mg cos θ = 0

Then collect like terms.

P (cos θ + μs sin θ) - mg (sin θ - μs cos θ) = 0

And this is just one step away from P = ...

 

[Jud]

Open the brackets.

P cos θ - mg sin θ + μs P sin θ + μs mg cos θ = 0

Then collect like terms.

P (cos θ + μs sin θ) - mg (sin θ - μs cos θ) = 0

And this is just one step away from P = ...

Sorry but I need bludgeoning in the head because I get 51.6N again, and I cannot see only one step available.

-P = (cos 30 + 0.4 sin 30) - 343.35 (sin 30 - 0.4 cos 30)
-P = -51.669N
P = 51.669N

 

[SammyS]

Homework Statement

[ IMG]http://questions.transtutors.com/Transtutors001/Images/Transtutors001_0f7b69b2-1e48-4ca4-8d27-89c487b0202a.PNG[/PLAIN]

Determine the minimum horizontal force P required to hold the crate from sliding down the plane. The crate has a mass of 35kg and the coefficient of static friction between the crate and the plane is μs = 0.40

2. The attempt at a solution
...

P cos 30 - 343.35 sin 30 + 0.40 (P sin 30 + 343.35 cos 30) = 0Now solving for P I get 51.6 N, which is incorrect.
The answer is given. P = 49.5 N

Can anyone give me an indication of where I have gone wrong please.

Thank You.

Solving

P cos 30° - 343.35 sin 30°+ 0.40 (P sin 30° + 343.35 cos 30°) = 0

for P gives 49.4689 .

Check your algebra.

 

[voko]

Sorry but I need bludgeoning in the head because I get 51.6N again, and I cannot see only one step available.

-P = (cos 30 + 0.4 sin 30) - 343.35 (sin 30 - 0.4 cos 30)

P (cos θ + μs sin θ) - mg (sin θ - μs cos θ) = 0 does not lead to that. P (cos θ + μs sin θ) means P multiplied by (cos θ + μs sin θ), not P plus (cos θ + μs sin θ).