Unformatted text preview: Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com ME 16 Spring 2009 Midterm 1 Instructions: 1. Do not open this booklet until instructed to do so. 2. Write your name on this answer booklet. 3. Read each question carefully. 4. If you ﬁnd that you are spending too much time on a problem, move on to the next and come back to it later. 5. Please, please, please write legibly. 6. To receive full credit you must show your work and explain clearly what you are doing. NAME: Solutions
With grading information Problem 1 24 points, Graded by: Rhoads Draw very carefully the Free Body Diagram for each of the following bodies. Your diagram should show all the forces acting on the speciﬁed body or bodies. Label the angles carefully. In each case, label the forces and explain in words what each are (e.g. Static friction, surface normal force, etc.) 1. Draw the FBD of the block. The block is sliding down the incline, and there’s no friction between it and the incline’s surface. Gravity points downwards. 2. Draw the FBD of the collar C. Gravity points downwards, and the collar C has non-zero mass. There is no friction between collar C and rod OA. applied to one of them. If the coeﬃcient of friction between any two surfaces is µ, answer the following two questions: a) What is 3. Draw the FBDs of all three be applied without m2 slipping over the maximum force which can bodies. Indicate carefully which forces are equal in magnitude m1 ? and the relevant directions. The coeﬃcient of friction between all surfaces is µ. b) For the force calculatedapplied force F is contact force between m1 and m3 ? Assume the in a) what is the insuﬃcient to cause movement. m2 F m1 m3 Solution: The maximum force on m2 is Ff r = µgm2 , therefore the maximum acceleration it can have is a = µg . a) The maximum total force is therefore F = a(m1 + m2 + m3 ) + (Ff r )all = = µg (m1 + m2 + m3 ) + µg (m1 + m2 + m3 ) = 2µg (m1 + m2 + m3 ) b) The contact force has to counteract the friction which block m3 feels and to provide the required acceleration of m3 , therefore Fcont = am3 + (Ff r )3 = am3 + µgm3 = 2µgm3 (15) Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
20 points, Graded by: Bamieh Produced with a Trial Version of PDF Annotator - www.PDFAnnotator.com
18 points, Graded by: Nowzari Problem 4 Graded by: Dandach A ball is placed on the pivoted platform shown at the instant when θ = 30o . There is no friction between the ball and platform. The ball is placed so that it is initially motionless relative to the ˙ platform. The platform is rotating at a constant θ > 0. Treat the ball as a point particle. ˙ How large should θ be so that ball moves upwards rather than downwards relative to platform? Hint: “Upward movement relative to platform” means that distance between pivot point and ball is increasing. 18 points, 20 points, Problem 5 Graded by: Dandach For each of the questions below, provide a brief (at most two sentences) answer. Solutions Manual Problem 2.8 vector a represents the acceleration of P ? (b)
(a) No, v is always tangent to the path. 1. Is it possible that (a) vector v represents the velocity of P ? (b) No, a must have a direction the between tangent and s it possible for the vector v shown to represent lyingvelocity of the point P ? E center of curvature. a = at + an , where at is tangent, and an points at center of curvature. lution 2.130 e vector v shown doesinﬂection points of the curve. path at point P . Hence v should not be considered E E Ignore not appear to be tangent to the velocity of P . o
is moving along the curve C , whose equation is given by y
2 2. A particle P moves along the curve C shown at constant speed v . What must be the angle φ 135 between the velocity vector v and the acceleration vector a. (Note that the angle φ may not be drawn accurately in the diagram) φ = 90 . x 2 Since v is tangent to path, and a = at . 2 .x an1is zero since D 4 xis constant. 2x ; /.2x 3/ speed 2 C y 2 speed vc . For any position on the curve C for which the radius is deﬁned (i.e., not equal to inﬁnity), what must be the angle 316 velocity vector v and the acceleration vector a. E E Problem 3.40
A plane is turning along a horizontal path at constant speed. What is the total force acting on the plane in the direction of the path? Solutions Manual onstant. This tells us that there is no component of acceleration in the direction of velocity. angle between v and a must be 90ı . E E
In the ut direction: ˆ 360 m at = 0. Ft = Note that v = constant Solutions Manual Problem 3.82 ⇒ at = 0. 4. The particle P is placed on the turntable and both are initially at rest. The turntable is then turned on so that the disk starts spinning. Assuming that there Solution is no friction between the turntable disk and the particle, what will be the motion of total force along the path is equal to zero because The the particle after the disk starts spinning? Explain. the tangential acceleration is equal to zero. The tangential acceleration is equal to zero because the speed of the plane is constant. The particle will stay where it was placed relative to an inertial frame (e.g. the room) since there Solution are zero lateral forces acting on P . The only forces acting on P will be gravity and a vertical surface normal force which cancels it.
The particle P will remain stationary relative to its inertial frame. There is no friction between the particle P and the turntable so when the turntable accelerates there will be no frictional force between P and the ...
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