Dynamics Exam1 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.

dynamics_1

 

 

 

 

 

 

 

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

dynamics_1solution

 

 

 

 

 

 

 

X component of force gives acceleration to the box.

FX=F.cos370=20.0,8=16N

FX=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)

 

dynamics_2

 

 

 

 

Free body diagram of these boxes given below.

dynamics_2solution

 

 

 

 

 

 

 

Components of force,

FX=F.cos370=30.0,8=24N

FY=F.sin370=30.0,6=18N

N1=m1.g-Fy=30-18=12N

N2=10N

Ff1 and Ff2 are the friction forces acting on boxes.

Ff1=k.N1=0,4.12=4,8N and Ff2=k.N2=0,4.10=4N

We apply Newton's second law on two boxes.

m1: Fnet=m.a

20-T-Ff1=3.a   20-T-4,8=3.a

m2: T-Ff2=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.

 

dynamics_3

 

 

 

 

Free body diagrams of boxes are given below;

dynamics_3solution


Fnet=m.a

(30-T)=2.5

T=20N


Fnet=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)

dynamics_4

 

 

 

 

 

 

 

 

 

 

Free body diagrams of boxes are given below;

dynamics_4solution

 

Since force acting on X is double of force acting on Y, aX=2aY

 

 

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.

dynamics_5

 

 

 

 

 

 

 

 

 

 

Free body diagrams of boxes are given below.

dynamics_5solution


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

 

 

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