Contact and electromagnetic force

[kuruman]

The contact force namely the resultant of friction and normal should be same and opposite to our weight. So 700N opposite of gravity.
We can also say the vector sum of gravity, normal and friction should be zero for the body to remain at rest.

Very good. You got it. There is only one contact force exerted by the incline and it has components like any other force. Its component perpendicular to the incline is also called the normal force and its component parallel to the incline is also called friction.

 

[rudransh verma]

Very good. You got it. There is only one contact force exerted by the incline and it has components like any other force. Its component perpendicular to the incline is also called the normal force and its component parallel to the incline is also called friction.

But why I got the wrong result. I broke the three forces in components and added the all vertical components. Similarly added all the horizontal ones. Got two eqns. Solved. But got Fn and f such that the resultant is some way more than 700N.

 

[kuruman]

But why I got the wrong result. I broke the three forces in components and added the all vertical components. Similarly added all the horizontal ones. Got two eqns. Solved. But got Fn and f such that the resultant is some way more than 700N.

You have posted here enough times to know that asking why you got the wrong result is insufficient. You have to show us what you did. Even though we often do a decent job as homework helpers, we suck consistently as mind readers.

 

[rudransh verma]

You have posted here enough times to know that asking why you got the wrong result is insufficient. You have to show us what you did. Even though we often do a decent job as homework helpers, we suck consistently as mind readers.

$F_N\cos78-f\cos12=0$
$F_N\sin78+f\sin12-700=0$
After solving,
$f=744 N$
$F_N=3638.5 N$

 

[haruspex]

$F_N\cos78-f\cos12=0$
$F_N\sin78+f\sin12-700=0$
After solving,
$f=744 N$
$F_N=3638.5 N$

Let's check if that satisfies your equations:
$3638.5\sin(78°)+744\sin(12°)=3556+155=3711$.
Not 700.
You could, and should, have done that check for yourself.
Please post your working if you cannot find your mistake.

 

[rudransh verma]

Let's check if that satisfies your equations:
$3638.5\sin(78°)+744\sin(12°)=3556+155=3711$.
Not 700.
You could, and should, have done that check for yourself.
Please post your working if you cannot find your mistake.

I got it! I was not multiplying by $\cos78$ to right hand side of second eqn.
I got $F_N=762.5N$
$f=155.5N$
$F_C=778.19N$ (maybe some small differences in decimal)

 

[haruspex]

I got it! I was not multiplying by $\cos78$ to right hand side of second eqn.
I got $F_N=762.5N$
$f=155.5N$

Does that satisfy your equations?

 

[rudransh verma]

Does that satisfy your equations?

No! But I know it’s because I have rounded in the middle of solution. Small difference of 6 N in first eqn.

 

[haruspex]

No! But I know it’s because I have rounded in the middle of solution. Small difference of 6 N in first eqn.

Your error is rather large, more than 10%. Please post your working.

 

[jbriggs444]

I got it! I was not multiplying by $\cos78$ to right hand side of second eqn.
I got $F_N=762.5N$
$f=155.5N$
$F_C=778.19N$ (maybe some small differences in decimal)

You are trying to come up with two components which combine (via the Pythagorean theorem) to produce a sum of 700 N.

It is immediately obvious that neither component can be greater than their vector sum.

 

[rudransh verma]

Please post your working.

$F_N\cos78-f\cos12=0$
$F_N\sin78+f\sin12=700$
So multiplying by sin78 in first and by cos78 in second eqn
$F_N\cos78\sin78-f\cos12\sin78=0$
$F_N\sin78\cos78+f\sin12cos78=700cos78$
On putting values of all trigonometric functions
$F_N(0.2)(0.97)-f(0.97)(0.97)=0$
$F_N(0.97)(0.2)+f(0.2)(0.2)=700(0.2)$

$f=142.72N$ and $F_N=692.19N$
Now its less than the resultant force.
But still there is a difference in the values.

 

[jbriggs444]

$F_N\cos 78-f\cos 12=0$
$F_N\sin 78+f\sin 12=700$
So multiplying by $\sin 78$ in first and by $\cos 78$ in second eqn
$F_N\cos 78 \sin 78-f \cos12 \sin 78=0$
$F_N\sin 78 \cos 78+f \sin 12 \cos 78=700 \cos 78$
On putting values of all trigonometric functions
$F_N(0.2)(0.97)-f(0.97)(0.97)=0$
$F_N(0.97)(0.2)+f(0.2)(0.2)=700(0.2)$

[after minor repairs on your $\LaTeX$]

Were you actually going to try to solve those equations?

The obvious next step would be to subtract the last two equations, thus arriving at an equation for $f$ alone. But if you were going to do that, it would be premature to substitute in values for the trig functions.

Although there are easier ways to break a vector into components, one can indeed do the job with simultaneous equations. Crucially, one must solve the simultaneous equations.

 

[kuruman]

Crucially, one must solve the simultaneous equations.

Indeed.
To @rudransh verma :
It would be best to not substitute numbers before solving because they often hide what is actually going and ho to take the next algebraic step. We have,
$F_N\sin\!\theta-f\cos\!\theta = 0~~~~~~~~~~(1)$
$F_N\cos\!\theta+f\sin\!\theta = W~~~~~~~~(2)$

I will follow your method and multiply the first equation by $\cos\!\theta$ and the second by $\sin\!\theta$
$F_N\sin\!\theta\cos\!\theta-f\cos^2\!\theta = 0~~~~~~~~~~~~~~~~~(3)$
$F_N\cos\!\theta\sin\!\theta+f\sin^2\!\theta = W\sin\!\theta~~~~~~~~(4)$

As @jbriggs444 suggested, subtract equation (3) from (4) to get
$f\sin^2\!\theta+f\cos^2\!\theta=W\sin\!\theta~\implies f=W\sin\!\theta$
Finally, put this value for $f$ in either (1) or (2) to get $F_N=W\cos\!\theta$,

You have been unable to get a reasonable answer. Why? My answer to that is that you have not yet acquired the single good habit that would limit algebraic and arithmetic mistakes and round off errors while simultaneously making it easy to troubleshoot your workings in case there is an actual mistake. That good habit is not substituting numbers until the very end.

What happened here? Not only you substituted numbers, but also you made things worse by mixing values of angles, using both 12° and 78°. I am sure you know that cos12° = sin78° but I am not sure that you would recognize on sight that (cos12°/sin78°) and (cos²12° + sin²78°) are both equal to 1.

I hope I have convinced you that your insistence on substituting numbers early on is actually a barrier to your ability to work out physics problems and a waste of your time. Using symbols instead of numbers makes it easier to be correct the first time around. I know that you been told that before, but here you have proof, three unsuccessful attempts at a solution, that your insistence on doing it your way does not work as well as you would want.