Dynamics Exam2 and Problem Solutions

Dynamics Exam1 and Problem Solutions

1. A box is pulled with 20N force. Mass of the box is 2kg and surface is frictionless. Find the acceleration of the box.

We show the forces acting on the box with following free body diagram.

X component of force gives acceleration to the box.

F_X=F.cos37⁰=20.0,8=16N

F_X=m.a

16N=2kg.a

a=8m/s

2. Picture given below shows the motion of two boxes under the effect of applied force. Friction constant between the surfaces is k=0,4. Find the acceleration of the boxes and tension on the rope. (g=10m/s², sin37⁰=0,6, cos37⁰=0,8)

Free body diagram of these boxes given below.

Components of force,

F_X=F.cos37⁰=30.0,8=24N

F_Y=F.sin37⁰=30.0,6=18N

N₁=m₁.g-Fy=30-18=12N

N₂=10N

F_f1 and F_f2 are the friction forces acting on boxes.

F_f1=k.N₁=0,4.12=4,8N and F_f2=k.N2=0,4.10=4N

We apply Newton's second law on two boxes.

m₁: F_net=m.a

20-T-F_f1=3.a 20-T-4,8=3.a

m₂: T-F_f2=1.a T-4=a

a=2,8m/s²

T=6,8N

3. As you can see in the picture given below, two boxes are placed on a frictionless surface. If the acceleration of the box X is 5m/s², find the acceleration of the box Y.

Free body diagrams of boxes are given below;

F_net=m.a

(30-T)=2.5

T=20N

F_net=m.a

T=5.a

20=5.a a=4m/s²

4. In the system given below ignore the friction and masses of the pulleys. If masses of X and Y are equal find the acceleration of the X?(g=10m/s²)

Free body diagrams of boxes are given below;

Since force acting on X is double of force acting on Y,a_X=2a_Y

For X: 2T-10m=m.a

For Y: T-10m=m.2a

a=2m/s²

5. When system is in motion, find the tension on the rope.

Free body diagrams of boxes are given below.

m₁: T+2g-20=2.a

m₂: 3g-T=3.a

5g-20=5.a

a=g-4 putting it into m₁ equation;

T+2g-20=2(g-4)

T=12N

Dynamics Exam2 and Problem Solutions

Dynamics Exam2 and Problem Solutions

1.Position time graphof the box is given below. Find the friction constant between box and surface? (g=10m/s²)

Slope of the graph gives us velocity of the box. Since the slope of the position time graph is constant, velocity of the box is also constant. As a result, acceleration of the box becomes zero.

F_net=F-f_s=m.a=0

F_net=f_s

f_s=12

k.mg=12

k.3.10=12

k=0,4

2. If the acceleration of the system given below is 3m/s2, find the friction constant between box and surface. (sin370=0,6, cos370=0,8, sin450=cos450=√2/2)

Free body diagrams of the system are given below.

Acceleration of the 10 kg box is 2m/s². Thus, net force acting on this box is;

F_net=m.a

F_net=10.2=20N

Date: 2016-04-22; view: 1289