Example (1): Motion of a block on a frictionless incline plane

Transcription

1 Firm knowledge of vector analysis and kinematics is essential to describe the dynamics of physical systems chosen for analysis through ewton s second law. Following problem solving strategy will allow you to tackle the problems with greater ease. Problem solving strategy: 1) Write the equation of motion F net = ma. Where, F net is the net force and a is the acceleration vector and is equal to a x i + a y j. 2) Draw a free body diagram (FBD) which shows all forces acting on the object/ objects. Do not draw the components of the force on the FBD. 3) Specify a coordinate system. Take x-axis in direction of the motion and y-axis perpendicular to the motion. 4) Express every force in a vector form. 5) In order to get the resultant force, F net, add all force vectors. 6) From the x and y components of F net, construct your x and y equations of motion. If the acceleration has a component in x direction, the x-component of the resultant vector is equal to ma x, otherwise it is equal to zero. Also, if the acceleration has a component in y direction, the y-component of the resultant vector is equal to ma y, otherwise it is equal to zero. 7) Depending on the question, solve the equation or the set of equations simultaneously for the unknown. Example (1): Motion of a block on a frictionless incline plane The figure below shows an object of mass m moving on a frictionless inclined plane which makes an angle with the horizontal. a) Find the acceleration of the block in terms of the constants g and. b) If the block started from rest and travelled a distance d, find the final speed of the block and the time it took to cover this distance. m

2 SOLUTIO: a) Our object is an accelerating block on a frictionless incline plane. ow, let s follow step-bystep the problem solving strategy and solve for the acceleration. STEP (1): Write the equation of motion F net = ma. Where, F net is the net force and a is the acceleration vector and is equal to a x i + a y j. STEP (2): Draw a free body diagram (FBD) which shows all forces acting on the object/ objects. Do not draw the components of the forces on the FBD. Based on our problem, we have only one object (the block) and there are two forces acting on the block. The forces are: 1) : Force of gravity due to the earth is pulling the block downward 2) : the normal force due to the surface is pushing the block perpendicular to the plane. STEP (3): Specify a coordinate system. Take x- axis in direction of the motion and y- axis perpendicular to the motion. The Fig. below shows the motion is in direction of x- axis while y-axis is perpendicular to the motion. The forces and acting on the block are also shown. y-axis a -i -j j i cos sin x-axis

3 STEP (4): Express every force in a vector form. This can be done by resolving every force vector into its components. Therefore, from the geometry of the above figure, vectors and can be written as = j (1) = sin i - cos j (2) STEP (5): In order to get the resultant force, F net, add all force vectors. Based on our problem, equations (1) and (2) can be added, we get F net = + = j + sin i - cos j Since F net = ma and a = a x i + a y j, we can write sin i + ( - cos) j = m (a x i + a y j) Where, (F net ) x = sin is the x-component of the resultant force (3) (F net ) y = - cos is the y-component of the resultant force (4) STEP (6): From the x and y component of F net, construct your x and y equations of motion. If the acceleration has a component in x direction, the x-component of the resultant vector is equal to ma x, otherwise it is equal to zero. Also, if the acceleration has a component in y direction, the y-component of the resultant vector is equal to ma y, otherwise it is equal to zero. Based on our problem, the motion of the block is in direction of x while there is no acceleration in y direction. Thus, from Eq. (3) and (4) the equations of motions are: sin = ma x and - cos = 0 STEP (7): Depending on the question, solve the set of equations simultaneously for the unknown. As we see, we were able to construct our equations of motion for the block. The acceleration can be found from the first equation, sin = ma x. The second equation has no

4 use in this problem unless we are asked to find the magnitude of the normal force. Thus, with = mg, we get mg sin = ma x. Therefore, a x = g sin Or in vector form a = g sin i ote, maximum acceleration is when = 90 o. This corresponds to a free falling object with a x = g = 9.8 m/s 2. Also, the acceleration does not depend on the mass of the object. b) Since the acceleration is constant, the final speed of the block can be found by using the kinematic equation, v 2 fx = v 2 ix + 2a x d. With v i = 0 and a x = g sin, we get v 2 fx = 2gd sin Therefore, v fx = 2gd sin The time it took the block to cover a distance d can be calculated using the kinematic equation, v fx = v ix + a x t With v ix = 0, v fx = 2gd sin and a x = g sin, we get 2gd sin = (g sin) t t = 2gd sin g sin t = 2d g sin

5 Example (2): Motion of a block on a rough incline plane The figure below shows an object of mass m moving on a rough inclined plane which makes an angle with the horizontal. If the coefficient of kinetic friction between the surface and the block is μ k, find the acceleration of the block in terms of the constants g, and μ k. m SOLUTIO: Problem solving strategy: STEP (1): F net = ma. Where, F net is the net force and a is the acceleration vector and is equal to a x i + a y j. STEP (2): Draw a free body diagram (FBD) which shows all forces acting on the object F f STEP (3) Specify a coordinate system and resolve the forces into x and y components. Take x- axis in direction of the motion and y-axis perpendicular to the motion. cos F f a y-axis sin x-axis -i -j j i

6 STEP (4): Express every force in a vector form. There are three forces acting on the block:, and F f. Therefore, from the geometry of the above figure, the three force vectors can be written as = j (1) = sin i - cos j (2) F f = F f i (3) STEP (5): In order to get the resultant force, F net, add all force vectors, Eq. 1, 2 and 3, F net = + + F f = j + ( sin i cos j ) F f i F net = ( sin F f ) i + ( cos) j (4) STEP (6): Construct your x and y equations of motion. The x-component of the resultant vector is equal to ma x and the y-component of the resultant vector is equal to ma y. Thus from Eq. (4), the equations of motion are: sin F f = ma x (5) cos = ma y = 0, since the acceleration in y direction is zero. (6) STEP (7): ow, let s solve for the acceleration. Since, F f = μ k = μ k cos, where we used Eq. (6) for the value of. Thus, Eq. (5) becomes, sin μ k cos = ma x With = mg, the magnitude of the acceleration is equal to a x = g (sin μ k cos) Or in vector form a = g (sin μ k cos) i

7 OTE: If the surface is frictionless, i.e. μ k = 0, the magnitude of the acceleration will be reduced to g sin as discussed in example number (1). Also note, the acceleration of the block on a rough plan is less than its acceleration on a smooth plane by the amount μ k g cos.