A block weighing 100 lbf sits on a 20º inclined surface. The coefficients of sta

Question

A block weighing 100 lbf sits on a 20º inclined surface. The coefficients of static and dynamic (kinetic) friction are 0.5 and 0.3 respectively. A force F = 10 t lbf is applied where t is in seconds.
(a) determine if the block will slide on the inclined surface when F = 0,
(b) find the time, t, when the force F begins to move the block up the incline,
(c) calculate the acceleration of the block when it just begins to move up the incline and (d) calculate the distance in feet the block moves in 20 seconds.

Explanation / Answer

First of all it is fun to see good engineering units again. I assume that the force F has units of force/time (a) Tan (theta) = u, Tan(20) = 0.36, so this block is not going to slide when F=0, but if it was moving, it would accelerate. (b) The force is not pushing up the incline. Part of the force is pushing the block into the incline F sin(30) or F/2, this will increase the frictional force. F cos(30) is directed straight up the incline. This force needs to overcome the component of the weight down the incline 100lbf sin(20) plus the frictional force N u. (Thanks for including gravity in this weight) The Normal force is 100lbf cos(20) + F/2 u is the poor man's mu; the coefficient of friction, to get it moving, we use 0.5 Putting it all together F up = 10tcos(30) F down = 100sin(20) + (10tsin(30)+100cos(20))0.5 when it just strats to budge these two forces are equal. Setting them equal and solving for t yields t = 13.17965s or 13.2 s (c) when the block just starts to move, the coefficient of friction switches over to kinetic. This gives the upward force an advantage. The component of F that is straight up the incline is still 114.1358 lbf, but now the downward force is only 82.162 lbf, so there are 31.97 lbf to accelerate the mass. I assume we divide by 32 ft/s^2 to get the mass. F = ma a = F/m 31.97/[100lbf/32 ft/s^2] = 10.32 ft/s^2 (d) this is unclear. Does the force continue to grow? Do the 20 seconds start at F = 0? Assuming the force stays constant at the value that caused the block to move and that time zero is when the force started, the acelleration will remain 10.32 ft/s^2 Then we can use distance = 1/2 a t^2 20s-13.17965022s = 6.820349777s d = 0.5 10.32 ft/s^2(6.82s)^2 = 238 ft This is the distance the block moves starting with F = 0 and the forces continuing at the value that first made the block slide.

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