Static Friction Problem - Physics I

[niravana21]

Static Friction Problem -- Physics I

Homework Statement

I've attached the problem, in order to avoid any confusion.2. The attempt at a solution

I tried to use F=ma for both of the masses, however, I need the acceleration along the x which I'm having diffuclity finding. Either that, or I taken a unnecessary route.

 

[Aaron K]

Since you are dealing with static friction, which is the amount of force you must overcome in order to move the object, you can set acceleration to zero in the problem leaving you with the mass of the block on the slope to solve for. In other words, the problem is asking "What is the smallest mass that can be placed on the slope before the system begins moving?"

 

[fisixC]

Calculate the sum of the forces on both objects seperatly, then solve for a common variable in one equation in order to substitute in anothe equation within the other object. Remembering that the force of friction is the coefficient of friction multiplied by the normal force of that object.

 

[niravana21]

Ok I can break down the sum of the forces on m1 as following:

Forces on m1 on the y = N - mgcos
Forces on m1 on the x = T - ukN-mgsin

do these components = zero or MA? if its MA, then I need the acceleration, which I am having trouble.

I tried solving, them setting them equal to zero, subbing in N=mgcos, and T=mg from the mass 2, but its not one of the options.

Any ideas now?

 

[Aaron K]

When setting acceleration to zero, ma also becomes zero for both masses. The reason you can set acceleration to zero is if you are working with static friction, there is no acceleration, once the system began moving you would be using kinetic friction.

Using the x direction for the m1 equation, T - (uk * N) - m1 * g * sin($$\theta$$) = 0, you can substitute in the value for tension from the sum of the forces in the y direction of m2 leaving m1 as the only unknown since if m1 is not moving, m2 cannot be moving, so you can say T = w2 or T = m2*g.

 

[niravana21]

what are you using to sub in the normal force for that equation?

I'm using N=mgcos, but that doesn't work?

 

[Aaron K]

That is what I used to solve the problem. Be sure you are using mass 1 for the mass in normal force. Solving for m1 will take some heavy algebra, or solver mode on a TI.

The equation should look like such when tension and normal force are substituted in:

m2*g - (m1*g*cos($$\theta$$)*$$\mu$$s) - (m1*g*sin($$\theta$$) = 0

 

[niravana21]

was your final answer 7.85?

 

[Aaron K]

No, using solver mode on a TI-89 I got 4.80121.

solve(0 = 5.5*9.81 - x*9.81*cos(23)*0.82 - x*9.81*sin(23),x) x here represents m1

 

[niravana21]

I forgot to multiply by "g" in one of them ahh. Haha thanks for the help

 

[Aaron K]

Not a problem.