
International Edition---engineering Mechanics: Statics, 4th Edition
4th Edition
ISBN: 9781305501607
Author: Andrew Pytel And Jaan Kiusalaas
Publisher: CENGAGE L
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Chapter 10, Problem 10.48P
To determine
The smallest value of parameter
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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 10 Solutions
International Edition---engineering Mechanics: Statics, 4th Edition
Ch. 10 - Determine the number of DOF for each of the...Ch. 10 - The uniform bar of weight W is held in equilibrium...Ch. 10 - Bars AB and AC of the mechanism are homogenous...Ch. 10 - The weight of each homogeneous bar of the linkage...Ch. 10 - The 1800-kg boat is suspended from two parallel...Ch. 10 - The 2.4-kg lamp, with center of gravity located at...Ch. 10 - The linkage is made of two homogenous bars of...Ch. 10 - For the frame shown, find the horizontal component...Ch. 10 - The four-bar linkage supports the homogeneous box...Ch. 10 - Prob. 10.10P
Ch. 10 - Determine the ratio P/Q of the forces that are...Ch. 10 - Find the vertical force P that will hold the...Ch. 10 - The linkage of the braking system consists of the...Ch. 10 - The automatic drilling robot must sustain a thrust...Ch. 10 - Determine the couple C for which the mechanism...Ch. 10 - The scissors jack is used to elevate the weight W....Ch. 10 - Prob. 10.17PCh. 10 - Calculate the torque C0 that must be applied to...Ch. 10 - Determine the force F and the angle a required to...Ch. 10 - Locate the instant center of rotation of bar AB...Ch. 10 - Prob. 10.21PCh. 10 - Determine the force P that will keep the mechanism...Ch. 10 - Prob. 10.23PCh. 10 - Prob. 10.24PCh. 10 - Prob. 10.25PCh. 10 - Determine the ratio P/Q for which the linkage will...Ch. 10 - Prob. 10.27PCh. 10 - Prob. 10.28PCh. 10 - If the input force to the compound lever is P = 30...Ch. 10 - Determine the roller reaction at F due to the...Ch. 10 - Prob. 10.31PCh. 10 - Prob. 10.32PCh. 10 - Prob. 10.33PCh. 10 - Prob. 10.34PCh. 10 - Prob. 10.35PCh. 10 - For the pliers shown, determine the relationship...Ch. 10 - When activated by the force P, the gripper cm a...Ch. 10 - Prob. 10.38PCh. 10 - The hinge is of the type used on some automobiles,...Ch. 10 - The spring attached to the sliding collar is...Ch. 10 - The weight W is suspended from end B of the...Ch. 10 - The uniform bar of weight W and length L = 1.8R...Ch. 10 - A slender homogeneous bar is bent into a right...Ch. 10 - The body shown is a composite of a hemisphere and...Ch. 10 - Prob. 10.45PCh. 10 - The uniform bar AB of weight W and length L is...Ch. 10 - Uniform rods of weights W1 and W2 are welded to...Ch. 10 - Prob. 10.48PCh. 10 - The semi-cylinder of radius r is placed on a...Ch. 10 - Prob. 10.50PCh. 10 - The spring attached to the homogenous bar of...Ch. 10 - The spring is connected to a rope that passes over...Ch. 10 - Find the equilibrium positions of the 30-lb...Ch. 10 - The mechanism of negligible weight supports the...Ch. 10 - Solve Prob. 10.54 assuming that A and B are...Ch. 10 - The stiffness of the ideal spring that is...Ch. 10 - Find the stable equilibrium position of the system...Ch. 10 - The uniform bar AB of weight W = kL is in...Ch. 10 - The weight of the uniform bar AB is W. The...Ch. 10 - The weightless bars AB and CE, together with the...Ch. 10 - Prob. 10.61PCh. 10 - The bar ABC is supported by three identical, ideal...
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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 findthe internal force andkeep either side of thearrow_forwardProcedure: 1- Cartesian system, 2(D)/(3)D, 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 \sum Fx=0 \sum Fy=0 \sum Fz=0 \sum Mx=0 \sum My=0 \Sigma Mz=0 2D \Sigma Fx=0 \Sigma Fy=0 \Sigma Mz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of the sectionarrow_forwardProcedure: 1- Cartesian system, 2(D)/(3)D, 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 \sum Fx=0 \sum Fy=0 \sum Fz=0 \sum Mx=0 \sum My=0 \Sigma Mz=0 2D \Sigma Fx=0 \Sigma Fy=0 \Sigma Mz=0 5- Use method of section and cut the element where you want to find the internal force and keep either side of the sectionarrow_forward
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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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