HW_Solns_5_part1 - 1 Physics 112 HW#5 Solutions Spring 2006 5-3 There is no acceleration in the y-direction so the sum of the forces in the

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: - 1 - Physics 112 - HW #5 Solutions Spring 2006 5-3 There is no acceleration in the y-direction so the sum of the forces in the vertical direction must vanish: W=2 T sin( θ ) (a) T = W 2 sin( θ ) = 9.8 N/kg × 90.0 kg 2 sin(10 ° ) = 2540 N (b) sin( θ )= W 2 T = 9.8 N/kg × 90.0 kg 2 × 2.5 ⋅ 10 4 N ⇒ θ = 1.01 ° 5-7 Write Newton's 2nd Law, Σ F = m a : +x: T B sin θ- T A = ma x = 0 +y: T B cos θ- mg = ma y = 0 (a) T B = mg cos θ = (4090 kg)(9.8 m/s 2 ) cos40° = 5.2 ∞ 10 4 N (b) T A = T B sin θ = mg cos θ sin θ = mg tan θ = (4090 kg)(9.8 m/s 2 ) tan40° = 3.4 ∞ 10 4 N 5-27 [Safe on Incline] Choose x-y axes parallel and perpendicular to the incline. Block is sliding at constant velocity, so sum of all forces on it is zero. Draw the FBD for the safe, and then apply N’s 2 nd Law parallel to and perpendicular to incline (let’s assume for now that we are pushing safe down the plane). Σ F x = W sin θ- F k + F = 0, where W = M g, so M g sin θ- F k + F = 0 (1) Σ F y = W cos θ- N = 0, so N =W cos θ (2) Block is sliding, so F k = μ k N Using (2), we have F k = μ k M g cos θ (3) Now look at Eq. (1): Mg sin θ- F k ± F = 0 If Mg sin θ > F k , then we need to pull on safe to keep it from accelerating down incline. If Mg sin θ < F k , then we need to push safe down incline. Using sin θ = 2.00 20.0 = 0.100 and cos θ = 0.995, find out whether M g sin θ- F k is +ve or –ve: Mg sin θ- F k = M g (sin θ- μ k cos θ ) = Mg (0.10 – (0.25)(0.995)) = Mg (-0.149) (4) So, the kinetic friction is larger than the component of the weight down the incline, so we need to push the safe to keep it moving down the plane....
View Full Document

This homework help was uploaded on 04/21/2008 for the course PHYS 1112 taught by Professor Leclair,a during the Fall '07 term at Cornell University (Engineering School).

Page1 / 8

HW_Solns_5_part1 - 1 Physics 112 HW#5 Solutions Spring 2006 5-3 There is no acceleration in the y-direction so the sum of the forces in the

This preview shows document pages 1 - 2. Sign up to view the full document.

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