Physics with Health Science Applications Ch. 3 pg. 56

Transcription

1 Physics with Health Science Applications Ch. 3 pg. 56 Questions 3.4 The plane is accelerating forward. The seat is connected to the plane and is accelerated forward. The back of the seat applies a forward force on you, accelerating you forward. Newton s third law of otion dictates that you will apply a force to the seat that is equal in agnitude and opposite in direction to the force that the seat applies to you. 3.8 Mass is a easure of an object s inertia-its tendency resist a change in its velocity. Mass is a property of an object and does not change with location. Weight is the force of gravity on an object. It can be calculated by the using the following forula: Wt = g where is the object s ass and g is the gravitational field strength and is equal to the acceleration of an object in free fall when air resistance is negligible. Because the value of g changes with location, an object s weight changes with location. 3.0 When an object oves in a curved path, its direction and therefore its velocity is changing. The only way to change an object s velocity is to apply an unbalance force to it. 3. In order for the clothes to go in a circle, an unbalance force ust be applied to the. The force is directed toward the center of the circle and is called a centripetal force. The barrel of the washing achine applies a centripetal force to the clothes but not to the. The goes out the holes in the wall of the barrel. centripetal force Probles pg. 58, Ch a = f f = a = 7kg 3.5 = 4.5 kg = 4.5 newtons 3. a= f = f 30 kg a =.5/sec = 10kg 3.3 a= f = x106 N 1. x10 6 kg = a) weight =g = 50kg( 9.8 b) wt =g = 50kg( 1.7 ) = 85 newtons kg ) = 490 = 490 newtons

2 Ch.3 pg. 58, Probles continued. 3.5 wt = g = wt g = 5x10 6 kg/ 9.8/ = 5.1x10 7 kg 3.6 See table 3.1, pg. 39 for the coefficient of friction of synovial fluid. f= µf N =.015(500 N)= 7.5 newtons. 3.7 f = µf N F N = f µ =.N.04 = 5newtons 3.8 f = µf N µ = f F N = 300n 1000n = 0.3 Check Table 3.1 for Steel on Steel dry 3.9 Tension in the thread = the weight of the spider. Tension tension= wt = g = 1 x 10-4 kg(9.8 / )= 9.8 x 10-4 newtons wt 3.11 f 90 kg 50N a = f = 50 n + f 90kg = 0.35 f = 90 kg(.35 ) 50n = 18.5newtons sec sec The negative sign indicates that the frictional force is backward. You should write your answer as f = 18.5 newtons backward.

3 Ch. 3 pg. 59 continued. 3.0 Resultant 150 Newtons We will use the Pythagorean theore to find the agnitude of the resultant. 00 Newtons c = a + b c = = 50 newtons is the agnitude of the resultant. The direction of the resultant can be found by using the trig. rule for the tangent of an angle. Tanθ = opposite adjacent = = 0.75 θ = 36.9 You should write your answer as 50 newtons 36.9 north of east or equivalently, 50 newtons at F 35 F 1 50 newtons Sin 35 = F 1 /50 newtons. Therefore, F 1 = Sin35 (50 N)= 8.7 newtons. Cos 35 = F / 50 newtons. Therefore, F = Cos 35 (50 N)= 41 newtons. 3.3 Step 1. Separate the 10 newton force into forward and left coponents. 10 newtons Sled Sin 30 (10 newtons)= 5 newtons 9 kg 30 to the left Cos 30 (10 newtons)= 8.66 newtons forward Step. Separate the 14.1 newton force into forward and right coponents. Sled 9 kg 45 cos 45 (14.1newtons)= 9.97 newtons forward Sin 45 (14.1newtons)= 9.97 newtons 14.1 n right.

4 Ch. 3 pg. 59 continued. Proble 3 continued. Step 3. Cobine forward forces and then cobine right-left forces newtons newtons = 18.6 newtons forward newtons - 5 newtons = 4.97 newtons to the right. Step 4. Deterine the su of the forward and right coponents newtons resultant = 4.97 newtons =19.3newtons. Tan ø = 4.97/18.6 =.67 ø= 15.0 Step 5. Now lets see what we ve got. The total force that the boys are applying is 19.3 newtons 15.0 to the right of forward. However, there is a frictional force back of 10 newtons. Therefore, the net force on the sled is = 9.3 newtons, 15.0 to the right of forward. The acceleration of the sled is found fro Newton s second law: a = f = 9.3n 9 kg = to the right of forward center of gravity F M c c- The weight of an object can be viewed as acting in a straight line fro the center of gravity Wt = g of the object, toward the center of the earth. F M is the unknown force produced by the splenius uscle. The ass of the head =5kg. Wt = g = 5 kg(9.8/ ) = 49 newtons We can solve this proble by using the concept that when a syste is in rotational equilibriu, the clockwise and counterclockwise torques are equal. Clock wise torque = Counter clock wise torque; F M = Therefore; 49n(.5 c ) 5c this anual. 49n(.5c) = F M (5c) = 4.5newtons. Which is a fairly large force. No wonder y neck aches as I type

5 Ch.3 pg F A F N = 784 newtons. I found this fro wt = g --4 c c = 80 kg(9.8/ )=784n Clockwise torque = Counter clockwise torque F A (4c)=784n(1c) F A = 784n(1c) = 350newtons 4 c 3.45 What acceleration, in /s, is experienced by aterials that are 10 c fro the center of rotation of a centrifuge that spins at 4,000 revolutions per inute? 1 First convert 10 c to eters: 10 ci 100c = V= distance tie = 4000 rev.( πi0.10 ) rev 60sec = 41.9 sec (41,9 To deterine the object s centripetal acceleration use: a = v 0.10 = 17, V= 35/sec An object that is oving in a circle, accelerates toward the center, F c A c = V centripetal acceleration r = 950 kg because there is a centripetal force F = a = 950 kg r = 00 eters (35 b) a = v 00 F C = A C acting on it. = 6.13 a) F = a = 950kg(6.13 sec sec ) = 580Newtons c) = 0.65 g

6 Chapter 3 continued ( a) b) a c = v.5 =19, V = 400k hr a = 400x sec =111 sec The equation for centripetal acceleration is a c = v r ; (111 a c = v 1000 = V= 45 rev 60sec ( π 1 rev )= 4.71 sec F a= v ( =. F=a= 1. kg(. / ) = 6.6 newtons 3.50 a=10g= 98 / a=v /r therefore a v= a r = 98 (5) = 49.5 sec sec To convert to revolutions per in. note that an object oving in a circle goes π r in 1 revolution and there are 60 seconds in 1 inute rev 60sec sec π 5 in =18.9 rev in