Enigma Posted June 12, 2005 Share Posted June 12, 2005 Could someone please help me with this question? I've had a go, but I don't really understand it. A block of mass 3kg is at rest on a rough horizontal plane. The block is acted on by a horizontal force of magnitude 14.7N. Given that the block is on the point of sliding, find the coefficient of friction between the block and the plane. Friction = coefficient of friction x normal reaction 14.7 = coefficient x (3 x 9.8) 14.7 = coefficient x 29.4 Coefficient of friction = 0.5 The horizontal force is now replaced by a force of magnitude P N acting downards at 30º to the horizontal. Given that the block is again on the point of sliding, find the value of P. The red is my working out - I'm not sure if it is correct. I don't know how to do the last part, so please could someone point me in the right direction. Many thanks, Enigma Link to comment Share on other sites More sharing options...
swansont Posted June 12, 2005 Share Posted June 12, 2005 If the force is now acting at an angle, there is an additional force down, meaning the normal force will be larger, since the acceleration is still zero. You will have to break P into its components. Link to comment Share on other sites More sharing options...
Enigma Posted June 13, 2005 Author Share Posted June 13, 2005 I have broken up P into its components, but I don't know where to go from there. Horizontal = P cos30 Vertical = P sin30 Does P cos30 = 14.7? Thanks, Enigma Link to comment Share on other sites More sharing options...
Enigma Posted June 13, 2005 Author Share Posted June 13, 2005 After some thought: R = 29.4 + P sin30 P cos30 = 0.5R P cos30/0.5 = R Substitute into top equation P cos30/0.5 = 29.4 + P sin30 Multiply right-hand-side to get rid of fracion P cos30 = 14.7 + 0.5(P sin30) P cos30 = 14.7 + 0.5p + 0.25 P cos30 - 0.5p = 14.95 P(cos30 - 0.5) = 14.95 P = 14.95/(cos30 - 0.5) P = 40.8N Is that correct? Thanks, Enigma Link to comment Share on other sites More sharing options...
mezarashi Posted June 13, 2005 Share Posted June 13, 2005 Pluggin in P = 40.8N doesn't seem to satisfy a zero net force condition. The question says on the verge of slipping, so in the x-direction, Force-friction and horizontal component of P are equal. Friction can only act parallel to a surface, normal forces perpendicular. Your freebody analysis should include the x and y axis. x-axis Pcos30 = Ffriction y-axis Psin30 + Fnormal = mg But we know that Ffriction = u x Fnormal, Fnormal = mg - Psin30 Try playing with those equations. Your answer shouldn't be more than 15N I'd suppose. Link to comment Share on other sites More sharing options...
Enigma Posted June 13, 2005 Author Share Posted June 13, 2005 OKay, I get this: P cos30 = 0.5R R = 29.4 - P sin30 P cos30 = 0.5(29.4 - P sin30) P cos 30 - 14.7 - 0.5P sin30 P cos 30 + 0.5P sin30 = 14.7 P(cos30 + 0.5P sin30) = 14.7 P = 13.2N But I don't understand how you got the equations out! Link to comment Share on other sites More sharing options...
mezarashi Posted June 13, 2005 Share Posted June 13, 2005 The equations are from the equilibrium condition. For example you know Newton's law: Fnet = ma If there is no acceleration (a), then there must be no net force. So we use this on the two axises, x and y. In the y direction, the forces are, the reaction force (Fnormal), gravitational force (mg), and the y-component of P (Psin30). In the x direction, again there is no acceleration so the force summation should be zero. The forces here are, Frictional Force (Ffriction) and the x-component of P (Pcos30). Take into consideration the directions and you will have the equations. Link to comment Share on other sites More sharing options...
Enigma Posted June 13, 2005 Author Share Posted June 13, 2005 Okay, I think I understand. I just hope I don't get a really hard question in the exam. Thank you very much Link to comment Share on other sites More sharing options...
Enigma Posted June 15, 2005 Author Share Posted June 15, 2005 I've managed to answer 3 parts of the following question, now I'm just stuck on the last part. Please can someone tell me how to go about this. Thanks. The displacement of a particle at time t seconds after it passes through a fixed point is s m, where s = 4.8t + 0.06t^2 - 0.004t^3 Write down expressions in terms of t for a) the velocity v = 4.8 + 0.12t - 0.012t^2 b) the acceleration a = 0.12 - 0.024t c) Find the value of s when the acceleration of the particle is zero. 0 = 0.12 - 0.024t 0.12/0.024 = t t = 5 s = (4.8 x 5) + (0.06 x 5^2) - (0.004 x 5^3) = 25.4 d) Find the distance travelled by the particle from the point where it reaches its maximum velocity to the point where its velocity is half its initial velocity. I'm thinking that when the particle has reached its maximum velocity, the acceleration will be zero, so the time will be 5 seconds, but I don't know how to finish the question. Many thanks, Enigma Link to comment Share on other sites More sharing options...
swansont Posted June 15, 2005 Share Posted June 15, 2005 I've managed to answer 3 parts of the following question' date=' now I'm just stuck on the last part. Please can someone tell me how to go about this. Thanks. The displacement of a particle at time t seconds after it passes through a fixed point is s m, where s = 4.8t + 0.06t^2 - 0.004t^3 Write down expressions in terms of t for a) the velocity [i']v = 4.8 + 0.12t - 0.012t^2[/i] b) the acceleration a = 0.12 - 0.024t c) Find the value of s when the acceleration of the particle is zero. 0 = 0.12 - 0.024t 0.12/0.024 = t t = 5 s = (4.8 x 5) + (0.06 x 5^2) - (0.004 x 5^3) = 25.4 d) Find the distance travelled by the particle from the point where it reaches its maximum velocity to the point where its velocity is half its initial velocity. I'm thinking that when the particle has reached its maximum velocity, the acceleration will be zero, so the time will be 5 seconds, but I don't know how to finish the question. Many thanks, Enigma You have an equation for v, so you can find the time when "its velocity is half its initial velocity." Now you have two times, and an equation that relates displacement and time. Link to comment Share on other sites More sharing options...
Enigma Posted June 16, 2005 Author Share Posted June 16, 2005 So can I find the initial velocity by putting zero into the velocity equation, and then find the time by using the initial velocity in the equation? Link to comment Share on other sites More sharing options...
mezarashi Posted June 16, 2005 Share Posted June 16, 2005 Yes, the velocity is at its maximum when the acceleration is zero. So you know that the maximum velocity occurs at t=5. A couple of other things after this. I'm assuming "distance travelled" really means what it means and not displacement from s=0. If that is the case. 1. Find the roots of the velocity equation and see if it the velecity reversed over the period t=0 to t=5. 2. If there are reversals, integrate the velocity in segments. Say from t=0 to t=3, then from t=3 to t=5. Take the absolute value of the answers and add them up (integrating velocity over time will give you distance). As for the second part of the question, initial velocity = 4.8, half of it is 2.4, so find the value for t and do the same test as stated above. Link to comment Share on other sites More sharing options...
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