Transcript Ch5prob

Chapter 5
TWO DIMENSIONAL FORCE PROBLEMS
Vehicle Motion with Friction
A box having a mass of 52 kg is placed onto an
incline of 35 o. The box experiences a friction force
of 45 N. a) Determine the μ for the surface. b)
Determine the direction and magnitude of Fnet . c) If
the box slides down the incline with acceleration,
determine the value of anet . The ramp is 10.0 m long
and the box starts from rest, determine the v f at the
bottom of the incline.
35 o
𝐹𝑤 = 𝑚𝑔 = (52)(−9.80) = −510𝑁
𝐹𝑤𝑥 = 𝑚𝑔sin35 = −510sin35 = −292𝑁
𝐹𝑤𝑦 = 𝑚𝑔cos35 = −510cos35 = −418𝑁
𝐹𝑓
45𝑁
𝜇= =
= 0.11
𝐹𝑛 418𝑁
𝐹𝑛 = −𝐹𝑤𝑦 = −(−418𝑁) = 418𝑁
𝐹𝑥 = 𝐹𝑤𝑥 + 𝐹𝑓 = −292𝑁 + 45𝑁 = −247𝑁
𝐹𝑥 −247𝑁
𝑎𝑥 = =
= −4.8 𝑚 2
𝑠
𝑚
52𝑘𝑔
𝑣𝑓 =
𝑣2
𝑜
+ 2𝑎𝛥𝑥 =
02 + 2(−4.8)(−10) = 9.8 𝑚
𝑠2
Sliding Box
At the request of your mother, you are asked to move a 50.0 kg box from the garage to
the kitchen. The box is too heavy for you to lift, so you slide it across the floor. You
apply a constant 525 N force onto the box from behind at an angle of 16 o which moves
the box with a constant velocity across the floor. a) Determine the Fn . b) Determine the
Ff . c) Determine the value for μ.
𝐹𝑤 = 𝑚𝑔 = (50)(−9.80) = −490𝑁
𝐹𝑎𝑦 = −525sin16 = −145𝑁
𝐹𝑎𝑥 = 525cos16 = 504𝑁
𝐹𝑓 = −𝐹𝑎𝑥 = −504𝑁
16 o
𝐹𝑦 = 𝐹𝑎𝑦 + 𝐹𝑛 + 𝐹𝑤 = 0; 𝐹𝑛 = −𝐹𝑤 − 𝐹𝑎𝑦 = −(−490) − (−145) = 635𝑁
𝜇=
𝐹𝑓 504𝑁
=
= 0.79
𝐹𝑛 635𝑁
Hanging Mass
A sign is suspended from two cables as is shown in the following diagram. The
mass of the sign is 150 kg. Determine the FT on each of the cables. The sign is
at rest both horizontally and vertically.
𝐹𝑤 = 𝑚𝑔 = (150)(−9.80) = −1470𝑁
𝑇𝑎𝑦 = 𝑇𝑎 sin55
𝑇𝑎𝑥 = −𝑇𝑎 cos55
𝑇𝐵𝑦 = 𝑇𝐵 sin35
𝑇𝐵𝑥 = 𝑇𝐵 cos35
𝐹𝑦 = 𝑇𝑎𝑦 + 𝑇𝐵𝑦 + 𝐹𝑤 = 0;
𝐹𝑥 = 𝑇𝑎𝑥 + 𝑇𝐵𝑥 = 0; 𝑇𝐵𝑥 = −𝑇𝑎𝑥
𝑇𝑎 cos55
𝑇𝐵 cos35 = −(−𝑇𝑎 cos55); 𝑇𝐵 =
; cos55 = sin35
cos35
𝑇𝑎 sin35
𝑇𝐵 =
; 𝑇𝐵 = 𝑇𝑎 tan35
cos35
𝐹𝑦 = 𝑇𝑎𝑦 + 𝑇𝐵𝑦 + 𝐹𝑤 = 0; 𝑇𝑎 𝑠𝑖𝑛55 + 𝑇𝑎 𝑡𝑎𝑛35 𝑠𝑖𝑛35 + −1470 = 0;
𝑇𝑎 = 1470 𝑠𝑖𝑛55 + 𝑡𝑎𝑛35𝑠𝑖𝑛35 = 1204 = 1200𝑁
𝑇𝐵 = 1200𝑡𝑎𝑛35 = 840𝑁
A
Sliding Car
The coefficient of friction between rubber tires and wet pavement is 0.50. The brakes are applied to a 750 kg
car traveling at 30 m/s which skids to a stop. a) What is the size and direction of the force of friction that the
road exerts on the car? b) What would be the size and direction of the acceleration as the car is stopping?
c) How far does the car travel before stopping?
𝐹𝑤 = 𝑚𝑔 = (750)(−9.80) = −7350𝑁
𝐹𝑛 = −𝐹𝑤 = −(−7350𝑁) = 7350𝑁
𝐹𝑓 = 𝜇𝐹𝑛 = (0.5)(7350𝑁) = −3675𝑁
𝐹𝑓 −3675
𝑎=
=
= −4.9 𝑚 2
𝑠
𝑚
750
𝑣𝑓2 − 𝑣𝑜2 02 − 302
𝛥𝑥 =
=
= 92𝑚
2𝑎
2(−4.9)
Atwood Machine Problem
Two masses are attached on either end of a weightless cord that is passed through a frictionless pulley. The
first mass is 5.0 kg and the second mass is 8.0 kg and both are free to move along the pulley. a) What is the
Fnet of the system? b) What is the anet of the system and its direction? c) What is the tension force found on
the cord? d) The second mass is 2.0 m above the floor when the system is started from rest. How long does it
take to reach the floor?
𝐹𝑊1 = 𝑚1 𝑔 = (8.0)(9.80) = 78.4𝑁
𝐹𝑊2 = 𝑚2 𝑔 = (5.0)(−9.80) = −49.0𝑁
𝐹𝑛𝑒𝑡 = 𝐹𝑊1 + 𝐹𝑊2 = 78.4 + (−49.0) = 29.4𝑁
𝐹𝑛𝑒𝑡
29.4
𝑎𝑛𝑒𝑡 =
=
= 2.26 𝑚 2
𝑠
𝑚1 + 𝑚2 8.0 + 5.0
𝐹𝑇 − 𝑚1 𝑔 = 𝑚1 𝑎1 ; 𝐹𝑇 = 𝑚1 (𝑎 + 𝑔) = (8.0)(2.26 + 9.8) = 96.5𝑁
𝑣𝑓 =
5.0 kg
m1
m2
8.0 kg
𝑣𝑜2 + 2𝑎𝛥𝑦 =
02 + 2(2.26)(2.0) = 3.0 𝑚 𝑠
𝑣𝑓 − 𝑣𝑜 3.0 − 0
𝑡=
=
= 1.33sec
𝑎
2.26
Projectile Motion
A metal ball rolls off of the edge of a table with a velocity of 2.5 m/s and reaches a point that is 1.50 m
away from the edge of the table. a) What is the total time of flight? b) How high is the table top above the
floor? c) What is the vertical velocity of the ball just prior to impact?
𝛥𝑥 1.50
𝑣𝑜𝑥 𝑡 = 𝛥𝑥; 𝑡 =
=
= 0.6sec
𝑣𝑜𝑥 2.50
1 2
1
𝛥𝑦 = 𝑣𝑜𝑦 𝑡 + 𝑔𝑡 = (0)(0.6) + (−9.80)(0.36) = −1.8𝑚
2
2
𝑣𝑓𝑦 =
𝑣𝑜2 + 2𝑔𝛥𝑦 =
02 + 2(−9.8)(−1.8) = 5.9 𝑚 𝑠
Projectile Motion
An artillery battery is set at the mouth of the Mississippi River to protect the country from a naval invasion from
England. The cannon has a 30o angle with the horizon and an elevation of 25.0 m above the river. The muzzle
velocity of the cannonball is 125 m/s. a) What is the time of flight for the cannon ball? b) What is the horizontal
range of the cannon? c) What is the horizontal velocity of the cannon ball? d) What is the vertical velocity of the
cannon ball?
𝑣𝑜𝑥 = 𝑣𝑜 cos30 = 125cos30 = 108 𝑚 𝑠 ; 𝑣𝑜𝑦 = 𝑣𝑜 sin30 = 125sin30 = 62.5 𝑚 𝑠
𝑣𝑓2 − 𝑣𝑜2 02 − 62.52
𝛥𝑦1 =
=
= 199𝑚
2𝑔
2(−9.8)
𝑣𝑓1 − 𝑣𝑜1 0 − 62.5
𝑡𝑢𝑝 =
=
= 6.38sec
𝑔
−9.8
𝛥𝑦𝑑𝑜𝑤𝑛 = 𝛥𝑦1 + 25 = 199 + 25 = 224𝑚
𝑣𝑓𝑑𝑜𝑤𝑛 =
𝑣𝑜2 + 2𝑔𝛥𝑦𝑑𝑜𝑤𝑛 =
𝑣𝑓𝑑𝑜𝑤𝑛 − 𝑣𝑜𝑑𝑜𝑤𝑛 −66.3 − 0
=
= 6.76sec
𝑔
−9.80
= 𝑡𝑢𝑝 + 𝑡𝑑𝑜𝑤𝑛 = 6.38𝑠 + 6.76𝑠 = 13.14sec
𝑡𝑑𝑜𝑤𝑛 =
𝑡𝑡𝑜𝑡𝑎𝑙
02 + 2(−9.8)(−224) = −66.3 𝑚 𝑠
Two Blocks and a Cord
A block of mass 5.00 kg rides on top of a second block of mass 10.0 kg. A person attaches a string to
the bottom block and pulls the system horizontally across a frictionless surface. Friction between the
two locks keeps the 5.00 kg block from slipping off. If the coefficient of static friction between the
two blocks is 0.350, what maximum force can be exerted by the string on the 10.0 kg block without
Fn
causing the 5.00 block to slip?
𝑇𝑜𝑝𝑏𝑙𝑜𝑐𝑘
𝐹𝑥 = −𝐹𝑓 = 𝑚𝑎; −𝜇𝐹𝑛 = 𝑚𝑎
-Ff
m
Ff
𝐹𝑦 = 𝐹𝑛 + 𝑚𝑔 = 0; 𝐹𝑛 = −𝑚𝑔; 𝜇(−𝑚𝑔) = 𝑚𝑎; −𝜇𝑔 = 𝑎
𝐵𝑜𝑡𝑡𝑜𝑚
M
mg
𝐹𝑥 = 𝑇 + 𝐹𝑓 = 𝑀𝑎; 𝑇 = 𝑀𝑎 − 𝐹𝑓 ;
𝑇 = 𝑀(−𝜇𝑔) + 𝑚𝑎; 𝑇 = 𝑀(−𝜇𝑔) + 𝑚(−𝜇𝑔)
𝑇 = −𝜇𝑔(𝑀 + 𝑚); 𝑇 = −(0.350)(−9.80)(10.0 + 5.0) = 51.5𝑁
Mg
T