{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

HW #5-solutions

# HW #5-solutions - wolz(cmw2833 HW#5 Antoniewicz(57420 This...

This preview shows pages 1–3. Sign up to view the full content.

wolz (cmw2833) – HW #5 – Antoniewicz – (57420) 1 This print-out should have 20 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001(part1of2)5.0points A 2 . 01 kg block is placed on top of a 8 . 17 kg block. A horizontal force of F = 63 N is applied to the 8 . 17 kg block, and the 2 . 01 kg block is tied to the wall. The coefficient of kinetic friction between all moving surfaces is 0 . 174. There is friction both between the masses and between the 8 . 17 kg block and the ground. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 01 kg 8 . 17 kg μ = 0 . 174 μ = 0 . 174 F T Determine the tension T in the string. Correct answer: 3 . 42745 N. Explanation: Given : m = 2 . 01 kg , M = 8 . 17 kg , m + M = 10 . 18 kg , μ = 0 . 174 , and F = 63 N . BasicConcept: Newton’s second law. Solution: Consider the free body diagram for the situation. m N 1 m g μ m g T M N 1 = m g N 2 M g μ ( m + M ) g μ m g F Applying Newton’s second law on block m yields m : summationdisplay F y = N 1 - m g = 0 (1) m : summationdisplay F x = f 1 - T = 0 , (2) where N 1 is the normal force exerted on block m by block M and f 1 is the frictional force between block m and block M . Using equa- tion (2) to solve for T and noting that from equation (1), N 1 = mg , we obtain T = f 1 = μ N 1 = μ m g = (0 . 174) (2 . 01 kg) (9 . 8 m / s 2 ) = 3 . 42745 N . 002(part2of2)5.0points Determine the magnitude of the acceleration of the 8 . 17 kg block. Correct answer: 5 . 1669 m / s 2 . Explanation: Applying Newton’s second law on block M yields M : summationdisplay F y = N 2 - N 1 - M g = 0 (3) M : summationdisplay F x = F - f 1 - f 2 = M a , (4) where N 2 is the normal force on block M by the ground and f 2 is the frictional force be- tween block M and the ground. From equa- tion (3), N 2 = M g + N 1 = M g + m g = ( M + m ) g . Thus f 2 = μ N 2 = μ ( M + m ) g Then solving for a from (4), we have a = 1 M ( F - f 1 - f 2 ) = 1 M bracketleftBig F - μ m g - μ ( M + m ) g bracketrightBig

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
wolz (cmw2833) – HW #5 – Antoniewicz – (57420) 2 = 1 M bracketleftBig F - μ (2 m + M ) g bracketrightBig = 1 8 . 17 kg braceleftBig 63 N - (0 . 174) bracketleftBig 2 (2 . 01 kg) + (8 . 17 kg) bracketrightBig × (9 . 8 m / s 2 ) bracerightBig = 5 . 1669 m / s 2 . 003(part1of2)5.0points A small metal ball is suspended from the ceil- ing by a thread of negligible mass. The ball is then set in motion in a horizontal circle so that the thread describes a cone. The acceleration of gravity is 9 . 8 m / s 2 . v 9 . 8 m / s 2 2 . 3 m 6 . 7 kg 20 What is the speed of the ball when it is in circular motion? Correct answer: 1 . 67508 m / s. Explanation: Let : = 2 . 3 m , θ = 20 , g = 9 . 8 m / s 2 , and m = 6 . 7 kg . Use the free body diagram below. T m g θ The tension on the string can be decom- posed into a vertical component which bal- ances the weight of the ball and a horizontal component which causes the centripetal ac- celeration, a centrip that keeps the ball on its horizontal circular path at radius r = sin θ . If T is the magnitude of the tension in the string, then T vertical = T cos θ = m g (1) and T horiz = m a centrip or T sin θ = m v 2 ball sin θ . (2) Solving (1) for T yields T = m g cos θ (3) and substituting (3) into (2) gives m g tan θ = m v 2 ball sin θ .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 9

HW #5-solutions - wolz(cmw2833 HW#5 Antoniewicz(57420 This...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online