Acceleration of a Single Body Sliding Down an Incline

Acceleration of a Single Body Sliding Down an Incline

Consider a mass “m” sliding down an incline and accelerating m v a Ɵ

Consider a mass “m” sliding down an incline and accelerating Draw an FBD m v a Ɵ

FN Draw an FBD fk m v a Fg Ɵ

FN Draw an FBD Set up an XY plane with positive x-axis in the direction of acceleration. fk m v a Fg Ɵ

FN Draw an FBD Set up an XY plane with positive x-axis in the direction of acceleration. fk m Y v a Fg Ɵ X

FN Draw an FBD Set up an XY plane with positive x-axis in the direction of acceleration. What real force is not lined up with the XY plane? fk m Y v a Fg Ɵ X

FN Draw an FBD Set up an XY plane with positive x-axis in the direction of acceleration. What real force is not lined up with the XY plane? Fg fk m Y v a Fg Ɵ X

FN Draw an FBD Set up an XY plane with positive x-axis in the direction of acceleration. It is easier to deal with forces lined up with the XY plane. How can we “correct” Fg to solve the conundrum? fk m Y v a Fg Ɵ X

FN Draw an FBD Set up an XY plane with positive x-axis in the direction of acceleration. It is easier to deal with forces lined up with the XY plane. How can we “correct” Fg to solve the conundrum? Resolve Fg into components. fk m Y Fgx v a Fg Fgy Ɵ X

FN Can you prove that Θ is also found in the little right triangle? fk m Y Fgx v a Fg Fgy Θ Ɵ X

FN Can you prove that Θ is also found in the little right triangle? The magnitude of the “weight” force Fg is “mg”. We will label this length on the little triangle. fk m Y Fgx v a Fg Fgy Θ Ɵ X

FN Can you prove that Θ is also found in the little right triangle? The magnitude of the “weight” force Fg is “mg”. We will label this length on the little triangle. fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = What is the force of gravity component parallel to the ramp? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = ? What is the force of gravity component perpendicular to the ramp? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp What is ay = ? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp ay = 0 fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp ay = 0 According to Newton’s first law, what is Fnet y = ? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp ay = 0 According to Newton’s first law, what is Fnet y = 0 fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 Vector Statement ? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement ? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement Scalar Statement ? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement FN - mg cos Ɵ = Scalar Statement fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement FN - mg cos Ɵ = Scalar Statement Therefore FN = ? fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement FN - mg cos Ɵ = Scalar Statement Therefore FN = mg cos Ɵ fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement FN - mg cos Ɵ = Scalar Statement Therefore FN = mg cos Ɵ label fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement FN - mg cos Ɵ = Scalar Statement Therefore FN = mg cos Ɵ mg cos Ɵ fk m Y Fgx v a Fg Fgy =mg Θ Ɵ X

FN Fgx = mg sin Ɵ the force of gravity parallel to the ramp Fgy = - mg cos Ɵ the force of gravity perpendicular to the ramp Fnet y = 0 FN + Fgy = 0 Vector Statement FN - mg cos Ɵ = Scalar Statement Therefore FN = mg cos Ɵ mg cos Ɵ fk m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN What is fk in terms of μK and FN ? mg cos Ɵ fk m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN mg cos Ɵ fk m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN What is fk in terms of μK , m , g , and Ɵ? mg cos Ɵ fk m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ mg cos Ɵ fk m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label mg cos Ɵ fk m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Let’s look at x-forces b/c the acceleration down the incline is along the x-axis. What is the equation of motion (Second Law equation) for this? mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation ? Vector statement mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector statement mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector statement ? Scalar Statement mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector statement mg sin Ɵ - μK mg cosƟ = max Scalar Statement mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector statement mg sin Ɵ - μK mg cosƟ = max Scalar Statement mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X What is the equation for the acceleration of a single body sliding down an incline?

FN fk = μKFN fk = μK mg cosƟ Label Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector statement mg sin Ɵ - μK mg cosƟ = max Scalar Statement mg cos Ɵ fk = - μK mg cosƟ m Y Fgx = mg sin Ɵ v a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X What is the equation for the acceleration of a single body sliding down an incline? ax = g sin Ɵ - μK g cosƟ ax = g (sin Ɵ – μKcosƟ) Memorize

Acceleration of a Single Body Sliding Up an Incline

Consider a mass “m” sliding up an incline and accelerating m v a Ɵ

Consider a mass “m” sliding down an incline and accelerating Draw an FBD m v a Ɵ

FN Draw an FBD m fk v a Fg Ɵ

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. ? Equation of motion or second law equation mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation ? Vector Statement mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector Statement mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector Statement ? Scalar Statement mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector Statement mg sin Ɵ + μK mg cosƟ = max Scalar Statement mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

What is the equation for the acceleration of a single body sliding up an incline? FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector Statement mg sin Ɵ + μK mg cosƟ = max Scalar Statement mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

What is the equation for the acceleration of a single body sliding up an incline? ax = g sin Ɵ + μK g cosƟ Memorize ax = g (sin Ɵ + μKcosƟ) FN Note that we still have the same components of Fg as before, the same FN value, and the magnitude of fk is the same. Fnetx = max Equation of motion or second law equation Fgx + fk = max Vector Statement mg sin Ɵ + μK mg cosƟ = max Scalar Statement mg cos Ɵ m Y Fgx = mg sin Ɵ v fk = + μK mg cosƟ a Fg =mg Fgy = - mg cos Ɵ Θ Ɵ X

Example: A block is released from rest on incline A 2
Example: A block is released from rest on incline A 2.00 m from the bottom of incline A. Incline A is inclined 36.9° as shown. The block slides down incline A onto a level, frictionless flat of length 3.00 m and then moves up incline B inclined at 53.1° as shown. The coefficient of kinetic friction between the block and incline A and B is Find the acceleration of the block as it slides… a) down incline A b) up incline B answer 4.40 m/s/s [down incline], 9.20 m/s/s [down incline] Find the total elapsed time for the block to slide from rest on incline A to maximum height along incline B at point F. Answer s s s = 2.12 seconds Incline B 2.00 m Incline A F 53.1° 36.9° 3.00 m Frictionless flat

Acceleration of a Single Body Sliding Down an Incline

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Acceleration of a Single Body Sliding Down an Incline

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