Statics practice midterm, Summaries of Mathematical Physics

Statics practice midterm with solutions

Typology: Summaries

2023/2024

Uploaded on 02/10/2025

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Queson 1 (20 points)
For the cables shown in figure 1, it is known that
the maximum allowable tension is 600 N in cable
AC and 750 N in cable BC. Determine:
(a) The maximum force P that can be applied at
C,
(b) The corresponding value of
.
Use the Trigonometric solution method
Figure 1
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QuesƟon 1 (20 points)

For the cables shown in figure 1, it is known that

the maximum allowable tension is 600 N in cable

AC and 750 N in cable BC. Determine:

(a) The maximum force P that can be applied at

C ,

(b) The corresponding value of .

Use the Trigonometric solution method Figure 1

Question 2 (30 points)

Pulleys A and B are mounted on bracket CDEF as shown

in Figure 2. The tension on each side of the two belts is as

shown.

  1. Determent the equivalent force-couple at F 2. Replace equivalent force-couple with a single

equivalent force, and

3. Determine where the single equivalent force line of

action intersects the bottom edge of the bracket

between points F and E. Figure 2

SOLUTION

Equivalent force-couple at A due to belts on pulley A

We have  F :  120 lb 160 lb  RA

R A  280 lb

We have  M (^) A : 40 lb(2 in.)  MA

M A  80 lb in.

Equivalent force-couple at B due to belts on pulley B

We have  F : (210 lb 150 lb) 25   R B

R (^) B  360 lb 25°

We have  M B :  60 lb(1.5 in.) MB

M B  90 lb in.

Equivalent force-couple at F

We have  F : R (^) F  ( 280 lb) j  (360 lb)(cos 25   i sin 25  j )

2 2

2 2

1

1

(326.27 lb) (127.857 lb)

350.43 lb

tan

tan

F

Fx Fy

Fy

Fx

R R

R R

R

R

i j

or R (^) FR 350 lb 21.4° ◄

We have  M F : MF   (280 lb)(6 in.)  80 lb in.

[(360 lb) cos 25 ](1.0 in.)

[(360 lb) sin 25 ](12 in.) 90 lb in.

 

   

M (^) F   (350.56 lb in.) k

To determine where a single resultant force will intersect line FE ,

350.56 lb in.

127 857 lb

2.7418 in.

F y

F

y

M dR

M

d R

 or d  2.74 in.◄