By A.F. Mills

The second one variation bargains entire assurance of warmth move with huge updated insurance that incorporates an emphasis on engineering relevance and on challenge fixing. Integrates software program to aid the reader in successfully calculations. conscientiously orders fabric to make textual content extra reader-friendly and obtainable. deals an intensive creation to warmth trade layout to reinforce the engineering and layout content material after all to fulfill ABET necessities. For pros in engineering fields.

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Additional info for Solution Manual for Heat Transfer

Sample text

A cannon ball is fired horizontally at a speed v0 from the edge of the top of a cliff of height H. e. the range) that the cannon ball travels. Check that your answer has the correct units. SOLUTION v0 H R In the x (horizontal) direction 1 x − x0 = v0x t + ax t2 2 Now R = x − x0 and ax = 0 and v0x = v0 giving R = v0 t. We obtain t from the y direction 1 y − y0 = v0y t + ay t2 2 Now y0 = 0, y = −H, v0y = 0, ay = −g giving 1 −H = − gt2 2 or t= 2H g Substuting we get R = v0 t = v0 2H g Check units: The units of v0 2H g are √ m −1 = m sec sec2 = m sec−1 sec = m m sec−2 which are the correct units for distance.

FORCE & MOTION - II Substitute for T and N into the left equation F cos θ − m2 a − m2 g − µ(m1 g − F sin θ) = m1 a F (cos θ + µ sin θ) − g(m2 + µm1 ) = m1 a + m2 a a= F (cos θ + µ sin θ) − g(m2 + µm1 ) m1 + m2 45 5. If you whirl an object of mass m at the end of a string in a vertical circle of radius R at constant speed v, derive a formula for the tension in the string at the top and bottom of the circle. SOLUTION T W R T W 46 CHAPTER 5. FORCE & MOTION - II Bottom: ΣFy = may mv 2 T −W = R T T mv 2 R mv 2 = mg + R = W+ Top: ΣFy = may mv 2 T +W = R mv 2 T = −W R mv 2 T = − mg R 47 6.

FORCE & MOTION - II 1. A mass m1 hangs vertically from a string connected to a ceiling. A second mass m2 hangs below m1 with m1 and m2 also connected by another string. Calculate the tension in each string. SOLUTION A) B) T’ T m m 1 2 T’ W m 2 2 Obviously T = W1 +W2 = (m1 +m2 )g. The forces on m2 are indicated in Figure B. Thus Fy = m2 a2y T − W2 = 0 T = W 2 = m2 g 39 2. What is the acceleration of a snow skier sliding down a frictionless ski slope of angle θ ? Check that your answer makes sense for θ = 0o and for θ = 90o .