Why are multiple resultant vectors possible for this problem?

[Muhammad Danish]

According to my understanding, option D is the only possible value of R. I don't understand how options A, B and C are included. Please explain this question.
Thanks.
(regards)

 

[drvrm]

According to my understanding, option D is the only possible value of R. I don't understand how options A, B and C are included. Please explain this question.
Thanks.
(regards)

just try to check the values of resultant at 0 and 90 degrees
anything outside this range of R is not possible.

 

[Muhammad Danish]

just try to check the values of resultant at 0 and 90 degrees
anything outside this range of R is not possible.

I do not understand...

 

[drvrm]

I do not understand...

in the question the angles are given between zero and ninety degree
so if you put zero
then the two vectors are collinear so they will add up that gives you a value of R
i think that is maximum value
similarly put them at 90 degree

 

[Muhammad Danish]

in the question the angles are given between zero and ninety degree
so if you put zero
then the two vectors are collinear so they will add up that gives you a value of R
i think that is maximum value
similarly put them at 90 degree

If we put them at 90 degrees, then the vectors will still add up to 7 N, won't they?

 

[Chestermiller]

If we put them at 90 degrees, then the vectors will still add up to 7 N, won't they?

Are you familiar with how to do vector addition?

 

[drvrm]

If we put them at 90 degrees, then the vectors will still add up to 7 N, won't they?

two vectors at an angle theta then there is a rule parallelogram -law
R = sqrt{ 3^2 + 4^2 + 2 (3*4 ) cos (theta) }
so if theta =90
cos 90 =0

 

[Muhammad Danish]

Are you familiar with how to do vector addition?

Yes..

 

[Chestermiller]

Yes..

Then you know that, unless they are co-linear and pointing in the same direction, their resultant magnitude can't be 7.

 

[Muhammad Danish]

two vectors at an angle theta then there is a rule parallelogram -law
R = sqrt{ 3^2 + 4^2 + 2 (3*4 ) cos (theta) }
so if theta =90
cos 90 =0

Oh I understand now, it means that if we put Cos 90, then the answer will be the lowest value. Similarly if we put Cos 0, then the outcome will be the maximum value. Maximum value=7 Minimum Value=5 so 4N is not possible. Am I correct?

 

[Muhammad Danish]

Then you know that, unless they are co-linear and pointing in the same direction, their resultant magnitude can't be 7.

When they will be co-linear, the resultant magnitude will be 7N, and when they are perpendicular to each other then the resultant magnitude will be 5N, so the value of R must lie between 5-7N. In the above MCQ the answer will be A.4N because it is not a possible value of R. Am I right?

 

[Chestermiller]

When they will be co-linear, the resultant magnitude will be 7N, and when they are perpendicular to each other then the resultant magnitude will be 5N, so the value of R must lie between 5-7N. In the above MCQ the answer will be A.4N because it is not a possible value of R. Am I right?

Yes.

 

[Muhammad Danish]

Yes.

Thanks Sir.

 

[Muhammad Danish]

two vectors at an angle theta then there is a rule parallelogram -law
R = sqrt{ 3^2 + 4^2 + 2 (3*4 ) cos (theta) }
so if theta =90
cos 90 =0

Thanks a lot! I understand now!

 

[haruspex]

if we put Cos 90, then the answer will be the lowest value

Within the range specified, yes, but more generally the lowest value will be when the vectors oppose each other, i.e. θ=180°, cos(θ)=-1.