test1.sol - N exerted on the skier at point A . Summing...

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

View Full Document Right Arrow Icon
EM 311 Test 1 Show all work . Each problem has equal value 1. A particle moves along the curve y = x 2 in the x - y plane. When x = 1, the x component of velocity of the particle is 2 m/s and the x component of acceleration is -1 m/s, determine a) the y components of velocity and acceleration; b) the unit tangent vector e t to the path; c) the radial and transverse components of velocity. a) The path is given by y = x 2 . Differentiating and using the chain rule, v y = dy dt = 2 x dx dt = 2(1)2 = 4 m/s. a y = d 2 y dt 2 = 2( dx dt ) 2 + 2 x d 2 x dt 2 = 2(2) 2 + 2(1)( - 1) = 6 m/s 2 . b) The tangent vector is given by e t = v || v || = 2 i + 4 j 20 . c) The polar angle θ = arctan( y/x ) = arctan(1 / 1) = π/ 4. The radial and transverse unit vectors are e r = cos θ i + sin θ j = i + j 2 e θ = - sin θ i + cos θ j = - i + j 2 Then v r = v · e r = (2 i + 4 j ) · ( i + j 2 ) = 4 . 24 m/s v θ = v · e θ = (2 i + 4 j ) · ( - i + j 2 ) = 1 . 41 m/s 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
2. The ski-jumper weighs 180 lb and attains a velocity of 80 feet/sec as he approaches takeoff. Calculate the magnitude of the normal force
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: N exerted on the skier at point A . Summing forces in the normal direction N-W cos 30 = m v 2 R Solving for N N = 180 32 . 2 80 2 150 + 180 cos 30 = 394 lb. 2 3. A ball of mass m = 2 kg is released from rest and falls 1 m before landing on a spring which is unstretched. After landing on the spring the ball continues to travel downward, compressing the spring .2 m before stopping. Determine the spring constant k . Using conservation of energy, m 2 v 2 1 + mgh 1 + 1 2 kS 2 1 = m 2 v 2 2 + mgh 2 + 1 2 kS 2 2 . Using the nal position as the datum point, h 1 = 1 . 2, h 2 = 0, S 1 = 0 and S 2 = . 2. The initial and nal velocities are zero, then solving for k , 2(9 . 81)(1 . 2) = 1 2 k ( . 2) 2 k = 1177 N/m. 3...
View Full Document

Page1 / 3

test1.sol - N exerted on the skier at point A . Summing...

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

View Full Document Right Arrow Icon
Ask a homework question - tutors are online