Chapter 5TWO DIMENSIONAL FORCE PROBLEMS

Vehicle Motion with Friction

35 o

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.

𝐹𝑤=𝑚𝑔=(52)(−9.80)=−510𝑁𝐹𝑤𝑥

=𝑚𝑔sin 35=−510 sin 35=−292𝑁𝐹𝑤𝑦

=𝑚𝑔 cos35=−510 cos 35=−418𝑁

𝐹 𝑛=−𝐹𝑤𝑦=−(−418𝑁 )=418𝑁

𝜇=𝐹 𝑓

𝐹 𝑛

= 45𝑁418𝑁

=0.11

∑ 𝐹 𝑥=𝐹𝑤𝑥+𝐹 𝑓=−292𝑁+45𝑁=−247𝑁

𝑎𝑥=𝐹 𝑥

𝑚=−247 𝑁

52𝑘𝑔=−4.8

𝑚𝑠2

¿

Sliding BoxAt 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.

16 o𝐹𝑤=𝑚𝑔=(50)(−9.80)=−490𝑁𝐹𝑎𝑦=−525 sin 16=−145𝑁𝐹𝑎𝑥=525 cos16=504𝑁𝐹 𝑓=−𝐹𝑎𝑥=−504𝑁

∑ 𝐹 𝑦=𝐹𝑎𝑦+𝐹𝑛+𝐹𝑤=0 ;𝐹𝑛=−𝐹𝑤−𝐹 𝑎𝑦=−(−490)−(−145)=635𝑁

𝜇=𝐹 𝑓

𝐹𝑛

=504 𝑁635𝑁

=0.79

Hanging Mass

A

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

𝑇 𝑎𝑦=𝑇 𝑎sin 55𝑇𝑎𝑥=−𝑇 𝑎cos 55𝑇 𝐵𝑦=𝑇 𝐵 sin 35𝑇 𝐵𝑥=𝑇 𝐵cos35

∑ 𝐹 𝑦=𝑇𝑎𝑦+𝑇 𝐵𝑦+𝐹𝑤=0 ;∑ 𝐹 𝑥=𝑇 𝑎𝑥+𝑇 𝐵𝑥=0 ;𝑇 𝐵𝑥=−𝑇 𝑎𝑥

𝑇 𝐵 cos35=−(−𝑇 𝑎 cos55);𝑇 𝐵=𝑇 𝑎 cos55

cos 35; cos55=sin 35

𝑇 𝐵=𝑇 𝑎sin 35

cos 35;𝑇 𝐵=𝑇 𝑎 tan 35

∑ 𝐹 𝑦  =𝑇 𝑎𝑦+𝑇 𝐵𝑦+𝐹𝑤=0 ;𝑇𝑎   𝑠𝑖𝑛55+(𝑇𝑎   𝑡𝑎𝑛35 )𝑠𝑖𝑛35+(−1470 )=0 ;

𝑇 𝑎=1470 /(𝑠𝑖𝑛55+𝑡𝑎𝑛35 𝑠𝑖𝑛35 )=1204=1200𝑁𝑇 𝐵=1200 𝑡𝑎𝑛35=840𝑁

Sliding CarThe 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)=−7350𝑁𝐹𝑛=−𝐹𝑤=−(−7350𝑁 )=7350𝑁

𝐹 𝑓=𝜇𝐹 𝑛=(0.5)(7350𝑁 )=−3675𝑁

𝑎=𝐹 𝑓

𝑚=−3675

750=−4.9

𝑚𝑠2

¿

Atwood Machine ProblemTwo 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?

m 1

m2

5.0 kg

8.0 kg

𝐹𝑊 1=𝑚1𝑔=(8.0)(9.80)=78.4 𝑁𝐹𝑊 2=𝑚2𝑔=(5.0)(−9.80)=−49.0𝑁

𝐹 𝑛𝑒𝑡=𝐹𝑊 1+𝐹𝑊 2=78.4+(−49.0)=29.4𝑁

𝑎𝑛𝑒𝑡=𝐹𝑛𝑒𝑡

𝑚1+𝑚2

=29.4

8.0+5.0=2.26

𝑚𝑠2

𝐹 𝑇−𝑚1𝑔=𝑚1𝑎1;𝐹 𝑇=𝑚1(𝑎+𝑔)=(8.0)(2.26+9.8)=96.5𝑁¿

Projectile MotionA 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?

𝑣𝑜𝑥𝑡=𝛥𝑥 ;𝑡=𝛥𝑥𝑣𝑜𝑥

=1.502.50

=0.6 sec

𝛥 𝑦=𝑣𝑜𝑦𝑡+12𝑔𝑡 2=(0)(0.6)+ 1

2(−9.80)(0.36)=−1.8𝑚

¿

Projectile MotionAn 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=𝑣𝑜cos 30=125 cos30=108𝑚𝑠;𝑣𝑜𝑦=𝑣𝑜 sin 30=125 sin 30=62.5

𝑚𝑠

¿

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 causing the 5.00 block to slip?

m

M

Mg

mg T

Ff

Fn

-Ff𝑇𝑜𝑝𝑏𝑙𝑜𝑐𝑘∑ 𝐹 𝑥=−𝐹 𝑓=𝑚𝑎;−𝜇 𝐹𝑛=𝑚𝑎

∑ 𝐹 𝑦=𝐹𝑛+𝑚𝑔=0 ;𝐹 𝑛=−𝑚𝑔 ;𝜇 (−𝑚𝑔)=𝑚𝑎 ;−𝜇𝑔=𝑎

¿𝐵𝑜𝑡𝑡𝑜𝑚∑ 𝐹 𝑥=𝑇 +𝐹 𝑓=𝑀𝑎;𝑇=𝑀𝑎−𝐹 𝑓 ;¿