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Unformatted text preview:  1  Physics 112  HW #5 Solutions Spring 2006 53 There is no acceleration in the ydirection 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 ° 57 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 527 [Safe on Incline] Choose xy 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....
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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).
 Fall '07
 LECLAIR,A
 mechanics, Acceleration, Force

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