PHYSICS 151 otes for Online Lecture. A free-bod diagra is a wa to represent all of the forces that act on a bod. A free-bod diagra akes solving ewton s second law for a given situation easier, because ou re odeling the sste as soething sipler than it actuall is. o draw a free-bod diagra: 1. Draw a separate diagra for each bod in the proble. If ou have to deal with two asses, for eaple, ou need two diagras.. Represent the bod as a point. or now, we don t know how to deal with forces that cause rotations. We re onl dealing with forces that cause linear otion. Regardless of what bod we re dealing with, we re going to represent it as a point. 3. Draw each of the forces that acts directl on the bod. here are two iportant ideas here: acts directl and on. Recall that onl forces that act directl on a bod are counted when ou su up the net force. he second operative word is on : ake sure that ou have included onl those forces acting on the bod. Don t include forces that the bod eerts on other things, or forces that aren t eerted on the bod that ou re analzing. 4. Draw all of the forces so that their tails are at the dot and their heads point in the direction of the force. 5. Label all of the forces E. 1: Draw a free-bod diagra for the eaple of a hand holding a block against the wall that is also suspended b a string. Solution: he block is represented as a point. We have four forces: tension due to the string, the applied force fro the hand, the weight of the block and the noral force of the wall on the block. he free-bod diagra looks like: A A g g 1
E. : A person is oving a 35 kg file cabinet b pushing it downward at an angle of 30 below the horizontal with an applied force of 35. What is the acceleration of the file cabinet? If ou draw a free-bod diagra for this situation, it will look like this: θ = 30 A W Part of the applied force acts to the right and part acts downward. he cabinet is accelerating to the right not up and down, so the net force on the cabinet ust be zero. hat eans that part of the applied force adds with the weight. he noral force has to support both the weight of the bo and the force downward that the person is appling. Regardless of the eact aount the applied force contributions downward, the noral force will alwas be greater than the weight. o he equation for ewton s second law in the -direction is = g Asin 30 = 0 which allows one to solve for the noral force as = g+ A sin 30 o. he iportance of the noral force will soon becoe apparent as we talk about friction. he equation for ewton s second law in the -direction is = Acos30 o A cos30 o = a for which one can solve for the acceleration a =. Σ = Acosθ = a A cosθ a = = = kg s ( 35 ) cos30 0.87 ( 35 ) You r How would the proble change if the force were applied 30 above the horizontal. (It wouldn t change our answer for acceleration at all in the present proble. It would change it quite a bit if we were considering friction because it would greatl affect the noral force.)
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E. 3 wo crewan pull a boat through a lock as shown. One crewan pulls with a force of 130 at an angle of 34 relative to the forward direction of the boat. he second crewan, on the opposite side of the lock, pulls at an angle of 45. With what force should the second crewan pull so that the net force of the two crewen is in the forward direction? 45 34 130 ( ) Σ = cos 45 130 sin 45 = a ( ) Σ = sin 45 130 sin 34 = 0 ( ) 130 sin 34 = = 103 s in 45 E. 4: An elevator is being planned for a building. If the elevator has a ass of 1.50 10 3 tension ust the supporting cable be able to withstand? Draw the forces acting on the elevator. kg, what + W = 0 = W = g = a = (1.50 10 3 kg)(9.80 s ) = 1.47 10 g 4 he elevator designer forgot that the plans specified that the elevator would eperience a aiu upward acceleration of 3.50 /s and installed a cable that could support 3.00 10 4 without breaking. Is the elevator safe? + 4
o deterine whether the elevator is safe, we need to find out what tension the cable will eperience when accelerating. If we re-draw our diagra, this tie including the acceleration in the picture, we find that when we appl ewton s second law, we have a slight change. = a W = a = W + a = g ( + a) = (1.5 10 3 kg) a = 3.5 /s ( 9.80 3.50 4 ) + =.00 10 s s 1.50 103 kg g + + You r What acceleration would the elevator have to eperience in order for the cable to break? Recall that the cable can handle 3.00 10 4. Answer: 10. /s You r What tension would the cable eperience if the elevator were being accelerated at 3.50 /s in the negative direction (i.e. down?) Hint: draw the diagra and ake sure ou get the right sign on the acceleration) Answer: 9.45 10 3 5
E. 5 A 65-kg skier speeds down a trail, as shown below. he surface is sooth and inclined at an angle of with the horizontal. (a) ind the direction and agnitude of the net force acting on the skier. (b) Does the net force eerted on the skier increase, decrease, or sta the sae as the slope becoes steeper? he net force on the skier results in the skier s otion. his otion is parallel to the slope, thus the direction of the net force is also parallel to the slope. W Σ = gcosθ = 0 = gcosθ Σ = gsinθ = a = gsin θ = (65 kg) 9.81 sin = 40 s You r A shopper pushes a 7.5-kg shopping cart up a 13 incline, as shown below. ind the agnitude of the horizontal force needed to give the cart an acceleration of 1.41 /s? = cosθ gsinθ = a (7.5 kg) 1.41 + ( 9.81 s s ) sin13 a ( + gsin θ ) = = = cosθ cos13 8 6