Hw10 sol - yet yet PDF

Preview

Summary

yet yet ...

Description

April 5, 2017

AAE 203 Spring 2017

Homework Ten Solution Exercise 1 Given: √ A force F¯ of magnitude F = 17 N as shown in Fig. 1.

Figure 1: Exercise 1 - problem setup. ¯ Q of F¯ about the point Q. Find: Moment M Solution: Since we are aiming to find the moment caused by F¯ about point Q, we must apply the ¯ = r¯ × F¯ . We begin by finding the force vector F¯ along the diagonal of the equation M rectangular prism from P to R. First, we must obtain the unit vector along the diagonal: x ¯P R 6ˆ e + 4ˆ e2 + 4ˆ e3 6ˆ e1 + 4ˆ e2 + 4ˆ e3 x ¯P R = 6ˆ e1 + 4ˆ e2 + 4ˆ e3 ⇒ fˆ = P R = √1 √ = . 2 2 2 k¯ x k 6 +4 +4 68 The force F¯ is found by multiplying the unit vector by the given force magnitude F = 50 N:  6ˆ √ e1 + 4ˆ e2 + 4ˆ e3  √ F¯ = 17 N = (3ˆ e1 + 2ˆ e2 + 2ˆ e3 )N. 68 Using r¯QP = −4 m eˆ3 , we have the moment as:   ¯ Q = r¯QP × F¯ = −4 m eˆ3 × F¯ = −4N m 3ˆ e2 − 2ˆ e1 + 0 M

 ¯ Q = 8ˆ M e1 − 12ˆ e2 Nm

1

Exercise 2 We are given a massless bar under dry friction (coefficient µ) at the ground as shown in Fig. 2. The roller support is still frictionless as in the previous exercise. Given:

Figure 2: Exercise 2 - problem setup. Find: Reaction on the bar at each support point. Solution: We proceed 2 and use the free body diagram (FBD) as shown in Fig. 3 P as in Exercise P ¯ ¯ ¯ to apply F = 0, M = ¯0 to the static problem. Note that the roller only gives a normal force, as there is no friction. Only the ground will generate a friction force. Except

Figure 3: Exercise 3 - FBD.  ¯ 2 = R2  − cos θˆ for the reaction force R e1 + sin θˆ e , the other forces are aligned with the 2 P ¯ = 0 off the FBD as: F reference frame axes. We may therefore read eˆ1 : Ff − R2 sin θ = 0

⇒ Ff = R2 sin θ

eˆ2 : R1 − F + R2 cos θ = 0 2

⇒ R1 = F − R2 cos θ

In order to throw out as many unknowns of the moment equation P as Apossible, we sum the moments around the contact point A with the ground to obtain M¯ = 0. Note that if we get the distance vector perpendicular to the force, the moment calculation r¯ × F¯ is simplified and the given F only acts with an effective distance of a cos θ which is equivalent to the distance from A to the intersection of vector F with the ground. a cos θ F ⇒ R2 = b  a cos θ sin θ a cos2 θ  ⇒ Ff = F F, R1 = 1 − b b

eˆ3 : R2 · b − F · a cos θ = 0

Plugging in values we obtain:  2 m · cos2 (65◦ )  2 m · cos (65◦ ) · 50 N · 50 N, R1 = 1 − 4m 4m 2 m · cos (65◦ ) · sin (65◦ ) Ff = · 50 N. 4m R2 =

Therefore, R1 = 45.54 N, R2 = 10.57 N, Ff = 9.58 N We also need to check whether the relationship holds that kFf k ≤ µkR1 k. Using the given friction coefficient we get: k9.58 Nk ≤ (0.5)k45.54 Nk k9.58 Nk ≤ 22.77 N Since this relationship holds, the bar is not moving, and the problem is static.

3

Exercise 3 Given: The force system as shown in the figure below.

Figure 4: Exercise 4 - problem setup. Find: The tension in each cable. Solution:

Figure 5: Exercise 4 - FBD.

We may split theP bar system up as shown in Fig. 5. Beginning with bar (I) in Fig. 5, the force balance F¯ = 0¯ is given as eˆ2 : T1 + T2 − W2 = 0 ⇒ T1 = W2 − T2 (bar 1). P A ¯ = 0 may be used to obtain another equation: The moment balance equation M eˆ3 : T2 · l2 − W2 · a2 = 0 ⇒ T2 = 4

a2 4m 80 N = 64N, W2 = 5 m l2

and hence, T1 may be solved for as T1 = W2 − T2 = 16 N. We may obtain an expression for the tension in cable 4 by applying

P

F¯ = 0¯ to bar (II):

eˆ2 : −(T1 + T2 ) − W1 + T3 + T4 = 0 ⇒ T4 = W2 + W1 − T3 (bar 2). Taking a sum of the moments about point B gives T3 · l1 − W1 · a1 − (T1 + T2 ) · b − T2 · l2 = 0, which allows us to solve for T3 as T3 =

W1 · a1 + (T1 + T2 )b + T2 l2 = 80 N. l1

Thus, T1 = 16 N, T2 = 64 N, T3 = 80 N, T4 = 60 N

5

Exercise 4 Given a uniform stepped cylinder with radii R = 1m, r = 0.5m and weight W = 100N is in static equilibrium on a rough horizontal plane and is attached to two cables at points A and B. Suppose µ = 1/4 and θ = 30◦ . Question: Find the maximum possible value of the mass m. (The gravity constant is g = 9.8m/s2 .)

Solution Figure 6 shows the force balance when the cylinder is considered the system. Consider the pulley system shown in the top right. We have T2 = mg

Figure 6: Cylinder FBD Here, it is noted that N is the normal reaction due to the cylinder being in contact with the surface and Ff ≤ µN is the friction force due to the fact that the surface is not perfectly smooth. Please note: The maximum possible value of the mass m will be find only when Ff = µN . The sum of forces is given by X F = −T1eˆ1 + Ff eˆ1 + T2 cos(θ)ˆ e1 + T2 sin(θ)ˆ e2 + N eˆ2 − W eˆ2 6

Since this is a statics problem which allows the above equation to be equated to zero. Therefore, the eˆ1 and eˆ2 terms of the sum of forces expression are separately equated to zero. First the eˆ2 direction is considered N − W + T2 sin(30o ) = 0 T2 →N =W− 2 Next the eˆ1 direction is considered. Here the expression for N derived in the previous expression is employed −T1 + µN + T2 cos(30o ) = 0  √  T2 3T2 → −T1 + µ W − + =0 2 2  √  T2 3T2 → T1 = µ W − + 2 2 Next the moments are summed about the center of mass of the stepped cylinder. This is done such that the sum of moments can be equated to zero. X M = −rT1eˆ3 + RµN eˆ3 − RT2 cos(θ)ˆ e3   T2 − RT2 cos(θ) = 0 → −rT1 + Rµ W − 2  √  1 3T2 T2 → −rT1 + − =0 W− 2 2 4 Plug in the expression for T1 in the previous step.   √  √  T2 3T2 T2 3T2 1 W− → −r(µ W − + − =0 )+ 4 2 2 2 2 200 → T2 = √ (12 3 + 1) Finally the above expression for T2 is substituted into T2 = mg. This gives the maximum value of mass m m=

200 √ 9.8 ∗ (12 3 + 1) → m = 0.937kg

Therefore the maximum mass is given as m = 0.937kg

7...