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COLLAPSE IMAGES Fx = 55 lb F-60 lb β=22 deg The magnitude of the force F....
P lb Q lb .5 0.5L F lb COLLAPSE IMAGES Q=192b P#112b F:53 lb h=7ft L=13ft...
P lb Q lb .5 0.5L F lb COLLAPSE IMAGES Q=192b P#112b F:53 lb h=7ft L=13ft α=27 deg β-80 deg The reaction at A. lb ENTER 3 tries remaining. 1 point(s) possible The vertical reaction at E lb ENTER 3 tries remaining. 1 point(s) possible The horizontal reaction at E NIR
Consider the following picture. goA W lb COLLAPSE IMAGES F2 = 31 lb W=93 b =...
Consider the following picture. goA W lb COLLAPSE IMAGES F2 = 31 lb W=93 b = 42 deg Find the magnitude of the force F, to keep the system in equilibrium. ENTER 3 tres remaining points) Sossible Find the direction of the force F, measured counter clockwise from the positive X-axis to keep the system in equilibrium deg ENTER stries remaining points possible
The hook and spring assembly shown in the figure are equilibrium, calculate: F. lb k lb/ft...
The hook and spring assembly shown in the figure are equilibrium, calculate: F. lb k lb/ft А (1) Falb В 02 2 ft F3 lb COLLAPSE IMAGES F1 = 110 lb F3 = 52 lb 01 - 21 deg 02 - 35 deg k = 1506 lb/ F2 = 98 lb The magnitude of the resultant force. Ib ENTER 3 tries remaining. 1 points) possible The orientation counter-clockwise from the positive x axis of the spring. deg ENTER 3 tries...
The x and z components of F are Fx and Fz. Calculate: Fx = 61 lb...
The x and z components of F are Fx and
Fz. Calculate:
Fx = 61 lb
Fz = 89 lb
β = 51 deg
1) The magnitude of the force F.
2) The magnitude of the y component of the force F.
nf
For the beam show, calculate: w lb/ft p lb/ft L ft aft bft COLLAPSE IMAGES w...
For the beam show, calculate: w lb/ft p lb/ft L ft aft bft COLLAPSE IMAGES w = 80 lb/ft p = 40 lb/ft L = 8 ft a = 8 ft b= 6 ft The vertical reaction at A. Іь ENTER 3 tries remaining. 1 point(3) possible The vertical reaction at B. lb ENTER 3 tries remaining. 1 point(3) possible The range for section 1. ) ft ENTER 3 tries remaining. 1 point(3) possible The shear force equation for section...
For the three force system shown, calculate: For the three force system shown, calculate: 1 F2...
For the three force system shown, calculate:
For the three force system shown, calculate: 1 F2 30 45° 3 COLLAPSE IMAGES F, 281 lb F 85 lb The magnitude of F2 such that the sum of F, F2, and Fs in the x direction is zero. 217.150 1 tries remaining 50 pointfs) possible b ENTER The direction of the resultant force FR deg ENTER 2 tries remaining. 50 pointis) possible
The plane ABC is hinged along BC. The force F is acting at 90° to the...
The plane ABC is hinged along BC. The force F is acting at 90° to the plane. Calculate: NG B(0, 0, Zb) ft F lb →Y C (0, Yc, 0) ft A(Xq, 0, 0) ft R + COLLAPSE IMAGES F = 278 lb Xa = 3 ft Zb= 8 ft Yc = 7 ft The force F in vector form. Oi + @j + OK ENTER 3 tries remaining. 1 point(s) possible The moment around B. Ci + Ok lb.ft...
: Given the system of forces below, calculate Given the system of forces below, calculate: F3...
:
Given the system of forces below, calculate
Given the system of forces below, calculate: F3 COLLAPSE IMAGES Write the F force in vector form k lb ENTER 3 tries remaining. 1 poinf(s) possible The resultant force vector of the system. k lb ENTER 3 tries remaining. 1 paint s) possible The magnitude of the resultant force. b ENTER 3 tries remaining. 1 paint's) possible The directional cosine alpha. ENTER 3 tries remaining. 1 paints) possible The directional cosine beta....
The truss is supported by wheels at A and a hinge at B and subjected to...
The truss is supported by wheels at A and a hinge at B and subjected to 2 equal vertical forces P lb at C and G, and a horizontal force P lb at D. Draw the reactions in the positive coordinate directions. P lb y L ft P lb D X A Lft L ft P lb COLLAPSE IMAGES P 175 lb L5 ft The x component of the reaction at A, Ax. ENTER 3 tries remaining 1 point(s) possible...
Q3 Weight: 20.00% Consider the beam below: w lb/ft Wlb h ft M lb-ft w ---HI...
Q3 Weight: 20.00% Consider the beam below: w lb/ft Wlb h ft M lb-ft w ---HI bft Cross section EE COLLAPSE IMAGES W = 62 lb/ft W= 139 lb /y = 7 ft 12 =2 ft b = 0.5ft h = 0.1 ft M = 127 lb.ft int.com/#!/assessment/2015/247/Homework%2013 The vertical reaction at A. lb ENTER 3 tries remaining. 1 points) possible The horizontal reaction at A 1/1 pts 0 lb System Answer 0 The vertical reaction at B. lb ENTER...
P lb Q lb .5 0.5L F lb COLLAPSE IMAGES Q=192b P#112b F:53 lb h=7ft L=13ft α=27 deg β-80 deg The reaction at A. lb ENTER 3 tries remaining. 1 point(s) possible The vertical reaction at E lb ENTER 3 tries remaining. 1 point(s) possible The horizontal reaction at E NIR
Consider the following picture. goA W lb COLLAPSE IMAGES F2 = 31 lb W=93 b = 42 deg Find the magnitude of the force F, to keep the system in equilibrium. ENTER 3 tres remaining points) Sossible Find the direction of the force F, measured counter clockwise from the positive X-axis to keep the system in equilibrium deg ENTER stries remaining points possible
The hook and spring assembly shown in the figure are equilibrium, calculate: F. lb k lb/ft А (1) Falb В 02 2 ft F3 lb COLLAPSE IMAGES F1 = 110 lb F3 = 52 lb 01 - 21 deg 02 - 35 deg k = 1506 lb/ F2 = 98 lb The magnitude of the resultant force. Ib ENTER 3 tries remaining. 1 points) possible The orientation counter-clockwise from the positive x axis of the spring. deg ENTER 3 tries...
The x and z components of F are Fx and
Fz. Calculate:
Fx = 61 lb
Fz = 89 lb
β = 51 deg
1) The magnitude of the force F.
2) The magnitude of the y component of the force F.
nf
For the beam show, calculate: w lb/ft p lb/ft L ft aft bft COLLAPSE IMAGES w = 80 lb/ft p = 40 lb/ft L = 8 ft a = 8 ft b= 6 ft The vertical reaction at A. Іь ENTER 3 tries remaining. 1 point(3) possible The vertical reaction at B. lb ENTER 3 tries remaining. 1 point(3) possible The range for section 1. ) ft ENTER 3 tries remaining. 1 point(3) possible The shear force equation for section...
For the three force system shown, calculate:
For the three force system shown, calculate: 1 F2 30 45° 3 COLLAPSE IMAGES F, 281 lb F 85 lb The magnitude of F2 such that the sum of F, F2, and Fs in the x direction is zero. 217.150 1 tries remaining 50 pointfs) possible b ENTER The direction of the resultant force FR deg ENTER 2 tries remaining. 50 pointis) possible
The plane ABC is hinged along BC. The force F is acting at 90° to the plane. Calculate: NG B(0, 0, Zb) ft F lb →Y C (0, Yc, 0) ft A(Xq, 0, 0) ft R + COLLAPSE IMAGES F = 278 lb Xa = 3 ft Zb= 8 ft Yc = 7 ft The force F in vector form. Oi + @j + OK ENTER 3 tries remaining. 1 point(s) possible The moment around B. Ci + Ok lb.ft...
:
Given the system of forces below, calculate
Given the system of forces below, calculate: F3 COLLAPSE IMAGES Write the F force in vector form k lb ENTER 3 tries remaining. 1 poinf(s) possible The resultant force vector of the system. k lb ENTER 3 tries remaining. 1 paint s) possible The magnitude of the resultant force. b ENTER 3 tries remaining. 1 paint's) possible The directional cosine alpha. ENTER 3 tries remaining. 1 paints) possible The directional cosine beta....
The truss is supported by wheels at A and a hinge at B and subjected to 2 equal vertical forces P lb at C and G, and a horizontal force P lb at D. Draw the reactions in the positive coordinate directions. P lb y L ft P lb D X A Lft L ft P lb COLLAPSE IMAGES P 175 lb L5 ft The x component of the reaction at A, Ax. ENTER 3 tries remaining 1 point(s) possible...
Q3 Weight: 20.00% Consider the beam below: w lb/ft Wlb h ft M lb-ft w ---HI bft Cross section EE COLLAPSE IMAGES W = 62 lb/ft W= 139 lb /y = 7 ft 12 =2 ft b = 0.5ft h = 0.1 ft M = 127 lb.ft int.com/#!/assessment/2015/247/Homework%2013 The vertical reaction at A. lb ENTER 3 tries remaining. 1 points) possible The horizontal reaction at A 1/1 pts 0 lb System Answer 0 The vertical reaction at B. lb ENTER...
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