Problem. Find the acceleration for the inclined plane problem with M 1 = 3kg

M 2 = 7kg

θ = 35!

µ = 0.1 (coefficient of friction)

Below is a detailed explanation of how I would find the answer to the problem. First I get into the problem by drawing a picture of it, labeling the masses and the angle of the incline. Then I choose an assumed direction of acceleration that both masses will experience. This step helps keeps the signs consistent throughout the problem. Concurrently, I need to identify all the forces acting so I sketch them on the diagram. I then split up the diagram to get the two free-body diagrams. To analyze the system, I need two sets of coordinates systems – one for each mass. I pick them so that one axis is in the direction of the assumed acceleration then the other coordinate has no acceleration associated with it. Once all that is done I write down Newton’s Second Law for each mass in its own coordinate system then solve the equations.

Diagrams – set up the problem 1) Draw a diagram of the problem. Note that I will put numbers in at the end so the solution is general until then

2) Choose a direction for the system acceleration. I will choose the acceleration such that M 1 system slides up the plane. If the acceleration turns out to be negative then M 1 is sliding down the plane.

Inclined Plane Analysis

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Fall 2014

3) Identify the forces acting on the masses.

4) Isolate the free body diagrams.

5) Choose the x and y axes for both masses. Chose one axis in the direction of assumed acceleration. Then there will be no acceleration in the other axis. For M 2 there is only vertical motion so no x axis is necessary.

Inclined Plane Analysis

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Fall 2014

Equations – Using Newton’s Second Law For each mass

!" " F = M a . i.e. ∑

∑F

x

= Max

and

∑F

y

= May

Mass 2 The tension, T, is in the - y direction. It’s pulling up on the mass. The weight, M 2 g , is in the + y direction. The net assumed acceleration is down.

W2 − T = M 2 a

Mass 1 First we have to decompose the weight into x and y components.

Wx = −W1 sin (θ )

x axis:

y axis:

Inclined Plane Analysis

Wy = −W1 cos (θ )

T − f − W1 sin (θ ) = M 1a T pulls to the right, f and Wx pull to the left. N − W1 cos (θ ) = 0 N pulls up, Wy pulls down.

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Collect Equations and Data Usually equations are scattered all over the paper so collect them all into one place. From Newton’s Second Law: W2 − T = M 2 a T − f − W1 sin (θ ) = M 1a N − W1 cos (θ ) = 0 Data: M 1 = 3kg M 2 = 7kg θ = 35! µ = 0.1 (coefficient of friction) We have (so far) three equations. We can have only three unknowns. Known M1 M2 θ µ (coefficient of friction)

Unknown T W2

a f W1 N

Since we have six unknowns, we need three more equations. They are 1) the weights in terms of masses W1 = M 1g W2 = M 2 g 2) an equation for the friction, f . f = µN

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Do the Math - the Physics is Done. We now have the three equations in three unknowns T, N, a: T − µ N − M 1gsin (θ ) = M 1a M 2 g − T = M 2a N = M 1g cos (θ ) The normal force, N is known: N = M 1g cos (θ ) We want to solve for a from the remaining two equations. T − µ N − M 1gsin (θ ) = M 1a M 2 g − T = M 2a

Since the tension, T, is an internal force, if you add the two equations, it will cancel. M 2 g − µ N − M 1gsin (θ ) = ( M 1 + M 2 ) a

Therefore the acceleration is M g − µ M 1g cos (θ ) − M 1gsin (θ ) a= 2 M1 + M 2 Let's put in the numbers now. 7 ⋅ 9.81− 0.1⋅ 3⋅ 9.81⋅ cos ( 35! ) − 3⋅ 9.81⋅sin ( 35 ) a= = 4.94 3+ 7 We are not yet through. Each variable has units. Since all entering units are kg, m, s (the SI system of units) the final units are Newtons and m/s2. Now that we have all the numbers, reread the problem statement to see which number it asks for. It wants the acceleration. Finally – the answer is m a = 4.94 2 s If the problem states to solve for the tension, T, use M 2 g − T = M 2 a . T = M 2 g − M 2a T = 7 ⋅ 9.81− 7 ⋅ 4.94 = 34.1 We know the unit is Newtons since tension is a force and we are using SI units. Therefore T = 34.1N Note the math is simpler if you solve for the acceleration then find the tension.

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Connection with Kinematics Sometimes, the problem asks for or gives the displacement or velocity or associated time. The connection between the Newton Second Law problem and the kinematic problem is the acceleration. You solve one of the problems for acceleration then plug the value of a in the other. For example, you are asked to find the tension in the rope in the above problem. You are given kinematic information, such as displacement, velocity, and/or time. Solve the kinematics problem for the acceleration. See my paper The Use of the Visual-Spatial Intelligence in the Solution of Elementary Physics Problems, available both on my web site’s main page, http://uhaweb.hartford.edu/ltownsend/, as well as in the physics section of the handouts page, http://uhaweb.hartford.edu/ltownsend/handouts.html#Physics. Once you have the acceleration, use the equations above to find the tension. T = M 2 g − M 2 a . On the other hand, you might be asked to find kinematic information given the above problem. Solve for the acceleration then use it in solving the kinematic equations for the desired variable.

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