System Dynamics
3rd Edition
ISBN: 9780073398068
Author: III William J. Palm
Publisher: MCG
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Chapter 3, Problem 3.26P
Consider the rack-and-pinion gear shown in Figure P3.20. Use the free-body diagram method to obtain the expression for the acceleration
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Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to find
Procedure:1- Cartesian system, 2D3D,type of support2- Free body diagram3 - Find the support reactions4- If you find a negativenumber then flip the force5- Find the internal force3D∑Fx=0∑Fy=0∑Fz=0∑Mx=0∑My=0\Sigma Mz=02D\Sigma Fx=0\Sigma Fy=0\Sigma Mz=05- Use method of sectionand cut the elementwhere you want to findthe internal force andkeep either side of the
Procedure:
1- Cartesian system, 2D3D,
type of support
2- Free body diagram
3 - Find the support reactions
4- If you find a negative
number then flip the force
5- Find the internal force
3D
∑Fx=0
∑Fy=0
∑Fz=0
∑Mx=0
∑My=0
ΣMz=0
2D
ΣFx=0
ΣFy=0
ΣMz=0
5- Use method of section
and cut the element
where you want to find
the internal force and
keep either side of the
Chapter 3 Solutions
System Dynamics
Ch. 3 - Prob. 3.1PCh. 3 - A baseball is thrown horizontally from the...Ch. 3 - For the mass shown in Figure 3.1.3b. m=10 kg, =25...Ch. 3 - A particle of mass m=19 kg slides down a...Ch. 3 - A particle of mass m slides down a frictionless...Ch. 3 - A radar tracks the flight of a projectile (see...Ch. 3 - Table 3.2.1 gives the inertia IO for a point mass...Ch. 3 - A motor supplies a moment M to the pulley of...Ch. 3 - Figure P3.9 shows an inverted pendulum. Obtain the...Ch. 3 - The two masses shown in Figure P3.10 are released...
Ch. 3 - The motor in Figure P3.11 lifts the mass mL by...Ch. 3 - Instead of using the system shown in Figure 3.2.6a...Ch. 3 - Consider the cart shown in Figure P3.13. Suppose...Ch. 3 - Consider the cart shown in Figure P3.13. Suppose...Ch. 3 - Consider the spur gears shown in Figure P3.15,...Ch. 3 - Consider the spur gears shown in Figure P3.15,...Ch. 3 - Derive the expression for the equivalent inertia...Ch. 3 - Prob. 3.18PCh. 3 - The geared system shown in Figure P3.19 represents...Ch. 3 - Prob. 3.20PCh. 3 - Prob. 3.21PCh. 3 - Prob. 3.22PCh. 3 - For the geared system shown in Figure P3.23,...Ch. 3 - For the geared system discussed in Problem 3.23,...Ch. 3 - The geared system shown in Figure P3.25 is similar...Ch. 3 - Consider the rack-and-pinion gear shown in Figure...Ch. 3 - The lead screw (also called a power screw or a...Ch. 3 - Prob. 3.29PCh. 3 - Derive the equation of motion of the block of mass...Ch. 3 - Assume the cylinder in Figure P3.31 rolls without...Ch. 3 - Prob. 3.33PCh. 3 - Prob. 3.34PCh. 3 - A slender rod 1.4 m long and of mass 20 kg is...Ch. 3 - Prob. 3.36PCh. 3 - Prob. 3.37PCh. 3 - The pendulum shown in Figure P3.38 consists of a...Ch. 3 - Prob. 3.39PCh. 3 - A single link of a robot arm is shown in Figure...Ch. 3 - 3.41 It is required to determine the maximum...Ch. 3 - Figure P3.42 illustrates a pendulum with a base...Ch. 3 - Figure P3.43 illustrates a pendulum with a base...Ch. 3 - 3.44 The overhead trolley shown in Figure P3.44 is...Ch. 3 - Prob. 3.45PCh. 3 - The “sky crane” shown on the text cover was a...
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- Design a 4-bar linkage to carry the body in Figure 1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis with the free choice values z = 1.075, q= 210°, ß2 = −27° for left side and s = 1.24, y= 74°, ½ = − 40° for right side. φ 1.236 P2 147.5° 210° 2.138 P1 Figure 1 Xarrow_forwardDesign a 4-bar linkage to carry the body in Figure 1 through the two positions P1 and P2 at the angles shown in the figure. Use analytical synthesis with the free choice values z = 1.075, q= 210°, B₂ = −27° for left side and s = 1.24, y= 74°, ½ = − 40° for right side. 1.236 P2 147.5° 210° P1 Figure 1 2.138 Xarrow_forwardcan you explain how in a coordinate frame transformation: v = {v_n}^T {n-hat} and then it was found that {n-hat} = [C]^T {b-hat} so v_n = {v_n}^T [C]^T {b-hat}, how does that equation go from that to this --> v_n = [C]^T v_barrow_forward
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