Link BD consists of a single bar 30 mm wide and 12 mm thick. Knowing that each pin has a10 mm diameter, determine the maximum value of the average normal stress in the link BD if θ = 15 degrees.

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Question:

Assume the following model of the economy, with the price level fixed at 1.0:

	C = 0.8(Y – T)	T = 1,000
	I = 800 – 20r	G = 1,000
	Y = C + I + G	Ms/P = Md/P = 0.4Y – 40r
	Ms = 1,200

A. Write a numerical formula for the IS curve, showing Y as a function of r alone. (Hint: Substitute out C, I, G, and T.)

B. Write a numerical formula for the LM curve, showing Y as a function of r alone. (Hint: Substitute out M/P.)

C. What are the short-run equilibrium values of Y, r, Y – T, C, I, private saving, public saving, and national saving? Check by ensuring that C + I + G = Y and national saving equals I.

D. Assume that G increases by 200. By how much will Y increase in short-run equilibrium? What is the government-purchases multiplier (the change in Y divided by the change in G)?

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Answer:

Solution

(a) 

AE=Y

Y= C+I+G

Y=0.8(Y-1000)+800-20r+1000

Y-0.8Y=1000-20r

0.2Y= 1000-20r

Y=5000-100r

(b) 

Ms/P=Md/P= 0.4Y-40r

12000/1=0.4Y-40r

Y=12000/0.4+40/0.4

Y=3000+100r

(c) 

IS=LM

5000-100r=3000+100r

2000=200r

r=10

Y=3000+100*10=4000

Y-T=4000-1000=3000

C=0.8(4000-1000)=2400

I=800-20*10=600

Sp=(Y-T)-C= 3000-2400=600

Sg= T-G= 1000-1000=0

S= Sp+Sg= 600+0= 600

We know

Y=C+I+G

Y-C-G=I and Y-C-G= S

So, S=I

(d)

G1=G0+200= 1000+200=1200

IS2:

Y=C+I+G

Y= 0.8(Y-1000)+800-20r+1200

Y= 0.8Y+1200-20r

(Y-0.8Y)=1200-20r

0.2Y=1200-20r

Y=6000-100r

LM1=LM0: Y=3000+100r

IS=LM

6000-100r= 3000+100r

3000=200r

r1=15

Y1= 4500

Y1-Y0= 

change in G, 

So,