EAS208_HW06_101810_Solutions

EAS208_HW06_101810_Solutions - EAS 208 (Fall 2010) Homework...

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EAS 208 (Fall 2010) Homework #6 – Solutions Problem 16–14 The box starts from rest and is subjected to the horizontal force described in the graph. The coefficients of friction between the box and the surface are 10 kg 0.2 sk μ == . Determine how fast the box is moving at 12 ts = . Answer: The box will not move until the force is able to overcome static friction. In the vertical direction ( F y - direction) there is no motion: 00 y FN W N W Σ=⇒−=⇒== m g The static friction is: 0.2 10 9.81 19.62 ss s f Nm g N = = 1
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The time that force overcomes static friction is: F 50 4 4 19.62 1.5696 45 0 5 0 ss s s s s F t ft s t = = ⇒= ⇒= ⋅ ⇒= s Apply the Linear Impulse – Momentum relation: I p = Δ r r Ff I Ip + r r r () 12 21 0 1 12 2 s s k Fd t tf t f mv mv ⎛⎞ +− = ⎜⎟ ⎝⎠ Note that is the area under the curve of 12 0 s t ( ) Ft during the time interval , and that the static friction is the same as the kinetic friction since 01 st s ≤≤ 2 0.2 sk μ == . Thus: 12 0 1 12 2 s s k t = 2 11 0 8 4 5 0 1 2 8 5 0 22 1 1.5696 19.62 12 1.5696 19.62 10 0 2 sN N s s N s k g v ⋅⋅ + − ⋅ +⋅ − ⋅ +− ⋅ = ⋅ − 2 10 180 m v s ⋅= 2 18 m v s = 2
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Problem 16–34 In a test of an energy-absorbing bumper, a car is driven into a barrier at 2800 lb 5 mi h . The duration of the impact is . When the car rebounds from the barrier, the magnitude of its velocity is 0.4sec 1.5 mi h . a) What is the magnitude of the average horizontal force exerted on the car during the impact? b) What is the average deceleration of the car during the impact? Answer: The velocities before and after the impact, in ft s , are: 1 5280 5 7.333 3600 mi ft h ft v hm i s s == 1 5280 1.5 2.200 3600 mi ft h ft v i s s a) Apply the Linear Impulse – Momentum relation for the duration of the impact: I p = Δ r r 2 1 21 t t Fd t mv mv =− r r r 3
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21 avg Ft m v m v Δ =− ( ) avg mv v F t + =− Δ ( ) 2800 2.200 7.333 0.4 avg Fl + b b 2072 avg = − The magnitude of the average horizontal force exerted on the car during the impact is . 2072 avg b b) The average deceleration of the car during the impact is: () ( ) 2 2.200 7.333 23.833 0.4 vv v ft a s tt −− Δ == = = ΔΔ 4
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Problem 16–52 A bullet with a mass of is moving horizontally with velocity and strikes a 5 block of wood, becoming embedded in it. After the impact, the bullet and block slide across the floor. The coefficient of kinetic friction between the block and the floor is 3.6 gr v kg 24 mm 0.4 k μ = . Determine the velocity v . Answer: During the impact the effect of the friction force can be neglected. Thus, we can assume that and we can apply Conservation of Linear Momentum: 0 ext I = r r 0 ext Ip p =Δ⇒Δ= r r r 12 0 p pp Δ =⇒ = r rr ( ) bullet bullet block mv m m v =+ r r () 21 bullet bullet block m vv mm = + r r bullet bullet block m = + r r 5
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After the impact we can apply the work-energy relation from position 2 (just after the impact) to position 3 (the block-bullet assembly comes to a rest): 12 UT = Δ () 0.024 22 32 0 11 dm k bullet block bullet block Nd s m m v m m v μ = −= + + ii There is o motion in the vertical direction ( y - direction), thus: ( ) 00 y bullet block FN W N W m m Σ=⇒−=⇒== + g
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This document was uploaded on 10/27/2011 for the course EAS 208 at SUNY Buffalo.

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EAS208_HW06_101810_Solutions - EAS 208 (Fall 2010) Homework...

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