The mass center of the 10-kg sliding door in Fig. (a) is located at G. The door is supported on the horizontal rail by sliders at A and B. The coefficient of static as well as kinetic friction is 0.4 at A and 0.3 at B. The door was at rest before the horizontal force P = 60 N was applied. (1) Find

Question

The mass center of the 10-kg sliding door in Fig. (a) is located at G. The door is supported on the horizontal rail by sliders at A and B. The coefficient of static as well as kinetic friction is 0.4 at A and 0.3 at B. The door was at rest before the horizontal force P = 60 N was applied. (1) Find the maximum value of h for which the door will slide to the right without tipping and the corresponding acceleration of the door. (2) If h = 1.5 m, find all forces acting on the door, and calculate its acceleration.

Accepted Answer

Part 1

The free-body and mass-acceleration diagrams for the door, shown in Fig. (b), are described in detail as follows.

Free-body diagram The FBD contains the following forces: the 10(9.8) = 98 N weight acting at G, the applied force P, the normal force [latex]N_A[/latex], and the friction force [latex]F_A = μ_AN_A = 0.4N_A[/latex]. Because the door starts from rest, its velocity will be directed to the right (i.e., in the same direction as the acceleration), which means that [latex]F_A[/latex] is directed to the left. There is no normal force and, therefore, no friction force at B, because the problem statement implies that the door is sliding to the right in a state of impending tipping about A.

Mass-acceleration diagram The MAD contains only the inertia vector of magnitude [latex]m\bar a[/latex] acting at G. There is no inertia couple because the door is translating (α = 0).

Inspection of Fig. (b) reveals that the unknowns are [latex]N_A[/latex] and h on the FBD, and [latex]\bar a[/latex] on the MAD. These three unknowns can be computed by deriving and solving any three independent equations of motion. When employing the FBD MAD technique, remember that (1) the resultant force on the FBD can be equated to the inertia vector [latex]m\bar a[/latex] on the MAD, and (2) the resultant moment about any point on the FBD can be equated to the resultant moment about the same point on the MAD.

Equating the x- and y-components of the forces on the FBD to the corresponding components of the inertia vector, we obtain

[latex]∑ F_y = m\bar a_y +\big\uparrow N_A − 98 = 0[/latex]
[latex]N_A[/latex] = 98 N

and

[latex]∑ F_x = m\bar a_x \underrightarrow{+} 60 − 0.4N_A[/latex] = 10a
60 − 0.4(98) = 10a

which gives

[latex]\bar a[/latex] = 2.08 m/s²

The third independent equation is a moment equation about any point. If we choose point A as the moment center, the resultant moment on the FBD is equated to the resultant moment on the MAD:

[latex](∑ M_A)_{FBD} = (∑ M_A)_{MAD} \circlearrowleft+ 60h − 10(9.8)(1.5) = 10\bar a(1.5)[/latex]

With [latex]\bar a[/latex] = 2.08 m/s², we get

h = 2.97 m

Part 2

The FBD and MAD for h = 1.5 m are shown in Fig. (c). The details of these diagrams are as follows.

Free-body diagram The FBD contains the 98-N weight, the applied force P = 60 N passing through G, the normal forces [latex]N_A[/latex] and [latex]N_B[/latex], and the friction forces [latex]F_A[/latex] and [latex]F_B[/latex]. From the solution of Part 1 we know that the door will be sliding to the right while maintaining contact at both A and B. The friction forces, determined by the kinetic coefficients of friction, are directed to the left—that is, opposite to the motion.

Mass-acceleration diagram Because the door is sliding to the right without rotating, the MAD contains only the inertia vector acting through G.

The number of unknowns in the FBD and MAD is three: [latex]N_A[/latex], [latex]N_B[/latex], and [latex]\bar a[/latex], which can be found from any three independent equations of motion. One such set of equations is given below; their validity can be determined by referring to the FBD and MAD in Fig. (c).

[latex]∑ F_y = m\bar a_y      +\big\uparrow    N_A + N_B − 98 = 0[/latex]           (a)
[latex]∑ M_G = 0      \circlearrowleft+      N_B(1) + 0.3N_B(1.5) − N_A(1) + 0.4N_A(1.5) = 0[/latex]           (b)
[latex]∑ F_x = m\bar a_x \underrightarrow{+}    60 − 0.4N_A − 0.3N_B = 10\bar a [/latex]           (c)

Solving Eqs. (a), (b), and (c) gives

[latex]N_A[/latex] = 76.85 N [latex]N_B[/latex] = 21.2 N [latex]\bar a[/latex] = 2.29 m/s²

The positive values of [latex]N_A[/latex] and [latex]N_B[/latex] confirm that the door will not tip.

Question 17.3: The mass center of the 10-kg sliding door in Fig. (a) is loc...

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