General Plane Motion, Forces & Accelerations; Prblm 16.1

“At a forward speed of 30 fps, the van brakes were applied, causing the wheels to stop rotating. It was observed that the van skidded to a stop at 20 ft. Determine the magnitude of the normal reaction and the friction force at each wheel as the van skidded to a stop.”

This was one (among of many) problems that I enjoyed whilst completing my dynamics course.

The first step is to create your FBD (Free Body Diagram) and set it equal to your KD (Kinetic Diagram). Because we’re dealing with an instant in which the van’s brakes shall be applied– the acceleration vector points to the left, or the –î (denoting deacceleartion). There is no motion whatsoever in the y-axis. So, our ma̅_y for this problem = 0.

Let’s analyze the FBD and KD

• ∑F_y^↑+ = ma̅_y = N_A + N_B – W = m(0) (i)
• ∑F_x^←+ = ma̅_x = f_A + f_B = ma̅_x (we know friction force, f = μ_kN) (ii)

The moment one here is a bit, tricky. We’ll need to solve for all five terms (in actuality it’s four– W, N_A, N_B, and the coefficient of kinetic friction, μ_k) eventually. Let’s take two moments. One about any wheel (let’s do B) and one about the center of mass gravity, G. It’ll make it easier to cancel out the terms later.

• ∑M_B^↻+ = I̅α + mad_⊥ (Any time you’re taking a moment abt. a point that’s not the center of mass gravity– you have to add this realignment factor “mad_⊥“, which is basically just the (acceleration scalar) × mass × the perpendicular distance from said non-C.O.G. point to G)
• ∑M_B^↻+ = N_A(12′) – W(7′) = (0) + -ma̅_x(4′) (van is not rotating, I̅α = 0)

Now for the moment about G

• ∑M_G^↻+ = I̅α = N_A(5′) – N_B(7′) + f_A(4′) + f_B(4′) = 0 (iii)
• eq. (iii) can be simplified to ∑M_G^↻+ = N_A(5′) – N_B(7′) + μ_kN_A(4′) + μ_kN_B(4′) = 0 (iv)
• Even furthur… N_A[5′ + μ_k(4′)] – N_B[7 – μ_k(4′)] = 0 (v)

Next, we must address the Work-Energy relationship expressed in the problem.

The work-energy equation is…

T₁ + V_p1 + V_s1 + U_1→2 = T₂ + V_p2 + V_s2

Where, 

• T = Kinetic Energy = ½mv²
• Vp = Potential Energy = mgh (N/A here)
• Vs = Spring Energy = ½kx² (N/A here)
• U_1→2 = Work Energy = Fx 

Filling in our two sceneraios (1 being the instance where the van slams the brakes, and 2 being a full stop at 20′ from braking point…)

½m(30 fps)² + [-f_A + f_B]20′ = ½m(0 fps)²

 = 0 (vi)

Now, we have at least 4-5 equations to play about with in terms of solving for unknown variables.

eq. (ii) and eq. (vi) both have the terms f_A, f_B, and m. Let’s try these two.

• (vi) f_A(20′) + f_B(20′) = ½m(900 ft²/s²) → f_A + f_B = $\frac{\frac{1}{2}m900{ft}^2{\mathrm{/s}}^2}{20\mathrm{‘}}$
• (vi) in (ii) = $\frac{\frac{1}{2}m900{ft}^2{\mathrm{/s}}^2}{20\mathrm{‘}}$ = ma̅_x (m term cancels, solve for a̅_x)
• a̅_x = [½⋅900 ft²/s²]/20ft = 22.5 fps² ←

We have a̅_x now. Let’s use the ∑M_B^↻+ equation to find the normal forces in the wheels.

• ∑M_B^↻+ = N_A(12′) – W(7′) = -m(22.5 fps²)(4′)
• simplified, this is = N_A(12′) – mg(7′) = -m(90 ft²/s²), {*g = 32.17 ft/s²}
• reordered to have two ‘m’ terms; → N_A(12′) – (225.19 ft²/s²)m = -(90 ft²/s²)m
• simplify again, we have N_A(12′) = (135.19 ft²/s²)m
• where, N_A = (11.266 fps²)m ↑ = 11.266m

We can plug this back into eq. (i), and solve for N_B

• eq. (i) says = (11.266 fps²)m + N_B – W = 0
• N_B = W – (11.266 fps²)m = mg – (11.266 fps²)m
• N_B = (32.17 fps²)m – (11.266 fps²)m
• N_B = (20.904 fps²)m ↑ = 20.904m

Now, we use eq. (ii) to solve for the μ_k

• eq. (ii) = μ_kN_A + μ_kN_B = m(22.5 fps²)
• μ_k[(11.266 fps²)m]+ μ_k[(20.904 fps²)m]= (22.5 fps²)m
• μ_k[(32.17 fps²)m] = (22.5 fps²)m
• μ_k = 0.6994 

Now, we may solve for the frictional forces, f_A and f_B

• f_A = μ_kN_A = (0.6994)(11.266 fps²)m = (7.879 fps²)m ←
• f_B = μ_kN_B = (0.6994)(20.904 fps²)m = (14.620 fps²)m ←

Now, the premis of the question says to find the reaction forces at each wheel. N_A = 2N_rear and N_B = N_front

• So, N_rear = (5.633 fps²)m ←
• So, N_front = (10.452 fps²)m ←