Question 1

resultant moment vector   at point O in Problem 2.2 is (in N &#8729; m) :

Accepted Answer

The resultant moment vector is calculated by summing the moments about point O. The given choice $12 j + 18 k$ matches the resultant moment vector calculated in the problem.

Question 2

Problem 2.7, the perpendicular distance d (in.) from force   to line AC is:&#10;(A) 1.7&#10;(B) 2.1&#10;(C) 1.2&#10;(D) 3.3

Accepted Answer

A

Question 3

Forces and couples are applied to ABCDE as shown in the figure. All turns of ABCDE are 90&#186; angles   . The statically equivalent moment at A (   for the force2couple system is (in N &#8729; m) :

Accepted Answer

C

Question 4

resultant force   (in newtons) for the force system shown in the figure is:

Accepted Answer

The answer of resultant force   (in newtons) for...

Question 5

Bracket ADO is acted on by forces F and T and couple M (see figure). Assume that F 5 8 N, T 5 9 N and M 5 12 N ∙ m. Joint coordinates are in meters.
The resultant force vector

(in newtons) is:

Accepted Answer

The answer of Bracket ADO is acted on by forces...

Question 6

Force   is applied at D along line DB and couple   is perpendicular to plane ABC. Assume magnitudes F 5 20 lb and couple M 5 50 in-lb. The moment

Accepted Answer

The answer of Force   is applied at D...

Question 7

Three couples of equal magnitude   e applied as shown. A single couple (in N &#8729; m) that is statically equivalent to the three couples is:     Equilibrium of rigid bodies; static friction

Accepted Answer

The answer of Three couples of equal magnitude  ...

Question 8

distributed loading on beam ABCD is to be replaced by a statically equivalent set of forces. The mag- nitude and location of the statically equivalent force on beam segment AB are:

(A) 575 lb, 4.1 ft
(B) 638 lb, 4.2 ft
(C) 740 lb, 3.9 ft
(D) 428 lb, 4.4 ft

Accepted Answer

To find the magnitude and location of the statically equivalent force on AB, we need to find the total load on AB, which can be found by taking the area under the distributed load curve:

Total load on AB = (2.5 + 1) * (300 + 500)/2 = 575 lb

Next, we need to find the location of the equivalent force. We can do this by finding the moment of the distributed load about any point on AB and setting it equal to the moment of the equivalent force about the same point. Let's take the moment about point A:

Moment of distributed load about A = (2.5/2) * 300 + (2.5 + 1/2) * 500 = 1375 lb-ft

Moment of equivalent force about A = F * 4.2 ft (assuming the force is applied 4.2 ft from A)

Setting these two moments equal, we can solve for F:

1375 lb-ft = F * 4.2 ft
F = 328.57 lb

Therefore, the magnitude and location of the statically equivalent force on AB are 328.57 lb and 4.2 ft, respectively. However, this answer is not one of the available choices. The closest available choice is B, which has a magnitude of 638 lb and a location of 4.2 ft. This suggests that there may be a mistake in the problem statement or the answer choices.

Question 9

Problem 2.7, the moment of force

Accepted Answer

To find the moment of force and couple about line AC, we need to project the force and couple vectors into a plane perpendicular to line AC, and then take the cross product of those projections with the unit vector along AC. 
Using the right-hand rule, we can see that the moment vector should point in the direction of positive AC (i.e. upwards). 
From the given vector components, we can see that only choice D has a positive component in the AC direction (24 k). Therefore, D is the correct choice.

Question 10

bracket ABCDEF, each y-direction force acts at the center of the bar segment to which it is applied (see figure). All turns of ABCDEF are 90º angles. The resultant moment at

for this parallel force system
Is (in N ∙ m) :

Accepted Answer

The answer of bracket ABCDEF, each y-direction force acts at...

Deck 2: Forces, Moments, Resultants