This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Solution for Practice Test 7 for Test 4 Solution to Practice Test Problem 7.1(Hospital ladder) Problem: In the intensive care unit at Arlington Memorial Hospital, the television sets are set in recessed areas high up the wall. Periodically, the 60kg Environmental Services worker must climb a ladder to reach the top of the TV to dust it. The 15kg ladder is 3m long. Because of equipment in the room suspended from the foot of the bed, the angle the ladder makes with the floor is 70 ◦ . The wall the ladder rests on is very slippery, but the floor is not. There is a coefficient of static friction between the ladder and the floor of 0.5. (a)Draw the extended freebody diagram for the ladder. For right now, just assume a position for the Environmental Services worker on the ladder (indicate it on your figure). (b)Write the equations for the horizontal ( bardbl ) and vertical ( ⊥ ) forces on the ladder. (c)Write the equation for the sum of the torques on the ladder. Solve your equations to find how high up the ladder the worker can go without the ladder slipping. You can solve for either the vertical distance or the distance along the ladder, just let us know which one! θ Solution to Part (a) m l = 15kg , ℓ = 3m , m ESw = 60kg , θ = 70 ◦ , μ s = 0 . 5 . ladder= l wall= w floor= f Environmental Services Worker= ESw Contact Force= F c Normal Force= F n Force of friction = F f 1 	 θ F n f,l F f f,l F w,l F g e,l F c ESw,l Grading Key: Part (a) 6 Points 1 point(s) : contact force due to worker straight downward at her position on the ladder and 4 times as big as gravity on ladder. 1 point(s) : Gravity straight downward and at center of ladder 1 point(s) : Forces balance. 1 point(s) : purely horizontal normal force due to the wall. 2 point(s) : force due to floor has (1pt) normal and (1pt) tangent compo nents (tangent component must be toward wall). Can be drawn as one force, with a slant toward the wall. Solution to Part (b) The ladder is not accelerating, so all forces must balance. To balance forces, we must look at each component individually. The sum of the horizontal forces is zero, and these are just the normal force due to the wall, and the 2 frictional force due to the floor. vector F f f,l + vector F n w,l = 0 vector F f f,l = − vector F n w,l Taking away from the wall to be positive F f f,l = F n w,l The sum of the vertical forces is also zero, vector F n f,l + vector F c ESw,l + vector F g e,l = 0 and taking upward to be positive, F n f,l = vector F c ESw,l + vector F g e,l We can actually solve this F n f,l = g ( m l + m ESw ) = 9 . 8m / s 2 (15kg + 60kg) = 735 . 75N If we plug this back into the horizontal forces equation F f f,l = F n w,l and use the fact that F f f,l = μ × F n f,l we get: F n w,l = μ × F n f,l = 0 . 5 × 735 . 75N = 367 . 88N ....
View
Full
Document
This note was uploaded on 05/04/2008 for the course PHYS 2054 taught by Professor Stewart during the Spring '08 term at Arkansas.
 Spring '08
 Stewart
 Physics

Click to edit the document details