Sol Idea: Assignment No.1 MCM517
Q.1 Answer
It is a fact that floating ads captures audience attention, advertisers can use eye catching imagery or visually engaging graphics and content for higher click through rates. This kind of advertisement have proven to increase the brand exposure. They can be seen by large number of people. Advertiser can also reach or target the specific audience for whom the ad has been positioned. They have proven to be more memorable and also have tracking capabilities.
But there are also drawbacks of this kind of online advertisement: as you may have also experienced that while watching a drama/movie/political talk show, we get irritated by the floating ads which requires minimum 5 seconds for skipping. The other drawback is that user can add ad blocking tools in their web browser to stop seeing the ads. But this kind of online advertisement have proven to increase the brand awareness and also the sales.
Q.2 Answer
Online journalist has to be very careful in gathering the information from sources. There are many problems that online journalist has to face in gathering and reporting the information like unreliable sources, inaccurate information, manipulation of data, fairness and balance etc. It has been seen that online reporters take less time to check their stories. They struggle with credibility and also objective reporting. Online journalist has to be more careful than journalist working in print and electronic media because of above mentioned problems.
Online journalist should not report the news story in a hurry just to get higher clicks and more traffic to their respective organization’s websites. The other big challenge is the role of gatekeepers; in online medium the role of responsible media professionals has become more important as it is disappearing day by day in online medium. The information is not filtered before it comes to public which has very bad effects on society.
SOLVED MTH304 Assignment 1 Solution and Discussion

Assignment # 1 MTH304 (Fall 2019)
Maximum Marks: 20
Due Date: 17 112019DON’T MISS THESE: Important instructions before attempting the solution of this assignment:
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• Also remember that you are supposed to submit your assignment in Word format any other like scan images etc will not be accepted and we will give zero marks correspond to these assignments.Question No. 1:
A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10
Question No. 2:
Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C. MARKS 10

Solution of Assignment # 1 MTH304 (Fall 2019)
Maximum Marks: 20
Due Date: 17 112019Question No. 1:
A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10
Solution
Question No. 2:
Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C. MARKS 10
Solution

Solution of Assignment # 1 MTH304 (Fall 2019)
Maximum Marks: 20
Due Date: 17 112019Question No. 1:
A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10
Solution
Question No. 2:
Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C. MARKS 10
Solution

@zareen said in MTH304 Assignment 1 Solution and Discussion:
Two particles of mass 3kg each are connected by a light inextensible string which passes over a smooth fixed pulley, which is attached to a string C. The string C is hanging on the fixed support. The particles are at rest. Find the tension in the string C.
Replace 6 with 3

@zareen said in MTH304 Assignment 1 Solution and Discussion:
A particle of mass 10kg is placed on an inclined plane which makes and angle of with the horizontal. Find the resolved parts of the weight of the particle in the direction parallel and perpendicular to the plane. MARKS 10
Free Body Diagram
T tension of string, W weight of the box, N force normal to and exerted by the inclined plane on the box, Fs is the force of frictionForces and their components on the xy system of axis.
Equilibrium: W + T + N + Fs = 0
Forces represented by their components
W = (Wx , Wy) = (  M g sin(35°) ,  M g cos(35°))
T = (Tx , Ty) = (T cos (25°) , T sin (25°) )
N = (0 , Ny) = (0 , N)
Fs = ( Fs , 0) = (  μs N , 0) , where μs is the coefficient of friction between the box and the inclined plane.
Sum of x components = 0 M g sin(35°) + T cos (25°) + 0  μs N = 0
which may be rewritten as
T cos (25°) = μs N + M g sin(35°)
sum of y components = 0 M g cos(35°) + T sin (25°) + N + 0 = 0
T sin (25°) = M g cos(35°)  N
We now need to solve the system of two equations with two unknowns T and N.
T cos (25°) = μs N + M g sin(35°) (equation 1)
T sin (25°) = M g cos(35°)  N (equation 2)
solve equation 2 above for N to get
N = M g cos(35°)  T sin (25°)
Substitute N by M g cos(35°)  T sin (25°) in eq 1 to get
T cos (25°) = μs [ M g cos(35°)  T sin (25°) ] + M g sin(35°)
rewrite above equation as follows
T [ cos (25°) + μs sin (25°) ] = μs M g cos(35°) + M g sin(35°)
Solve for T
T =
 M g cos(35°) + T sin (25°) + N + 0 = 0
μs M g cos(35°) + M g sin(35°)
cos (25°) + μs sin (25°)
Substitute with numerical Values
μs = 0.3, M = 10 Kg, g = 10 m/s^2
T ≈ 79.3 NUse N = M g cos(35°)  T sin (25°) found above
N = 100 cos(35°)  79.3 sin (25°) ≈ 48.4 N  M g sin(35°) + T cos (25°) + 0  μs N = 0