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Problem Set 6: Static Equilibrium and Torque, Work-Kinetic Energy ...

Problem Set 6: Static Equilibrium and Torque, Work-Kinetic Energy ...

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Once the puck starts to slide as the ramp’s angle is increased, read this value for theta <strong>and</strong>plug into equation (5).Another way is to connect a spring gauge to the end of a block of material you are tryingto find the coefficient of friction for.mgFmFfNFigure 4: Spring gauge system to determine the coefficient of friction.Summing the forces in the x <strong>and</strong> y directions produces a relationship between thecoefficient of friction, weight, <strong>and</strong> the force applied.∑ Fx= F − f(6)=> F = µ N∑ Fy= N−mg(7)=> N = mgFµ = (8)mgOnce the block starts to move as a force is applied to it, read the value of the force off thespring gauge. This value can be plugged into equation (8) to find the coefficient offriction.PROBLEM 3: How much work does it take to slide a hockey puck up the ramp?


YXFfriction_rampFpushθmgNFigure 5: Pushing a hockey puck up an inclined plane.Only forces in the x-direction are considered.<strong>Work</strong> = Force X distance.18∑ Fx = FPush −mgsinθ −Ffriction_ramp(9)From the y direction the Normal force, N is found to meN = mg cosθ(10)Thus the forces in the x-direction are∑ Fx= FPush−mgsinθ −µ mg cosθ(11)=> F − mg +Push( sinθ µ cosθ)The total distance to travel is determined from Figure 6.


DYθXFigure 6: Variables of ramp.Where X <strong>and</strong> θ are given dimensions on the table (reference 2.007 web site). This makesthe value of D asDX= (12)cosθSubstituting this back into the work equation produces the final result of the amount ofwork done pushing a puck up the ramp.( Push ( sinθ µ cosθ))W = F − mg +Push( tanθ µ )=> F X − mg + XXcosθ(13)PROBLEM 4: How much work does it take to roll a hockey puck up the ramp?Ffriction_pusherYXFpushFfriction_groundmgNFigure 7: FBD of a hockey puck being rolled up an incline.


∑ Fx = FPush − mgsinθ + Ffriction_ramp(14)From the y direction the Normal force, N is found to meN = mg + F(15)cosθfriction_pushWhere F friction_push = µ Push F Push . Substituting this value for the one in (14), the normalforce from the ramp becomes:N = mg + F(16)cosθ µPush PushThe forces in the x-direction are:∑( cos)F = F −mgsinθ − µ mg θ + µ Fx Push ramp Push Push( 1 µ µ ) mg ( sinθ µ cosθ)=> F − − +Push ramp Push rampSubstituting this back into the work equation produces the final result for the amount ofwork done rolling a puck up the ramp.( Push( 1 µrampµ Push) ( sinθ µrampcosθ))W = F − − mg +FPushX=> − − +cosθ( 1 µrampµ Push) mg( tanθ µramp )XXcosθPROBLEM 5: Does it really matter, given the work required to increase the hockeypuck’s potential energy, whether you slide or roll the puck up the ramp?Potential energy in this situation is equal to(17)(18)U= mghThe value of h is found from Figure 6, which ish=X tanθThus the potential energy isU = mgX tanθ


The potential energy is the same for both cases, but the work is different because onehockey puck is rolling, less work, <strong>and</strong> the other is sliding, more work. Refer to equations(13) <strong>and</strong> (18).PROBLEM 6: How else might you get the hockey puck (s!) up the ramp quickly, <strong>and</strong>what would be the forces <strong>and</strong> powers involve?!a) Carry the pucks on the machine <strong>and</strong> travel up the ramp. This removes the frictionforce due to the puck rolling against the car, but the additive force is the frictionbetween the wheels <strong>and</strong> ground <strong>and</strong> the puck remaining on the back of the car.This is like problem set 3.( µ )W = F − N dcar( µcar ( car puck( s) ) cosθ)=> F − m + m g d(19)Power is work divided by time. Thus guestimating the amount of time it would take todrive the car up the ramp will give us the amount of power.Power<strong>Work</strong>= (20)timeb) Catapult/launch projectile.βFigure 8: Launch hockey pucks instead of push or roll them.FmtgD= (21)sin2βD is the length of the ramp as in Figure 6. <strong>Work</strong> is the force times the total distance theprojectile is shot, which in our case the distance is X in Figure 6. This produces


WmtgDXsin2β= (22)Power is also equal to the force times the velocity. In our case the velocity is the changein distance divided by time, which produces:Powerm gD X= (23)tsin2βtimePROBLEM 7: Guestimate what you think is a reasonable time for the motion of thehockey puck, given the total time for the contest, <strong>and</strong> now estimate what is the powerrequired for the different strategies.The Equation of power is given in (20). Guessing it will take a time T to travel up theramp, this produces the following power for sliding <strong>and</strong> rolling the hockey puck:Sliding:PowerRolling:( tanθ µ )F X − mg + XPush= (24)TPushPower= F X ( 1−µ rampµ Push) − mg ( tanθ + µramp )X(25)T cosθTNote, the student may pick any value for T given that the amount of time considers thetime it takes it gets to the ramp <strong>and</strong> the total length of the contest of 45 seconds.PROBLEM 8: Remember the platter with the balls <strong>and</strong> shot-put? How much energymust be expended to get it spinning to the point where the hockey balls fly off? Howmuch energy must be expended to get the shot-puts to fly off?From <strong>Problem</strong> <strong>Set</strong> 5, we foundax2rh−hω = + gR Rr ( − h)2(26)We also know the velocity of the platter is ωR eff <strong>and</strong> the kinetic energy (KE) is

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