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Qmechanic ♦
Qmechanic ♦
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Psuedo Pseudo force as seen by aan accelerating frame of refrence

A question struck my mind when i was trying to solve the following problem,   

Problem statement

I was able to solve it by just considering forces in the horizontal and vertical direction however the solution turned out to be very lengthy, i found a solution for it online which used psuedo/fictious forces to solve the problem, the solution was as follows

Suppose there is an observer on the wedge and the wedge is accelerating towards the left with acceleration $a$. Then the psuedo force on the block of mass $m$ is $ma$.

Balancing forces on $m$ normal to the plane we get $$N + ma\sin\theta = mg\cos\theta$$ Balancing forces on $M$ in $X$ direction $$F + N\sin\theta = F\cos\theta + Ma$$

Solving these two equations gives me the solution to the above question as follows

$$a = \frac{mg\sin\theta\cos\theta + F(1-\cos\theta)}{M+m\sin^2\theta}$$

This does match with my answer and with the answer in my textbook, however i do not understand why there would be a psuedo force on the wedge of mass $M$, amounting to $Ma$.

With respect to a observer on the wedge should there be no psuedo force on it?.

Any help would be highly appreciated!

Psuedo force as seen by a accelerating frame of refrence

A question struck my mind when i was trying to solve the following problem,  Problem statement

I was able to solve it by just considering forces in the horizontal and vertical direction however the solution turned out to be very lengthy, i found a solution for it online which used psuedo/fictious forces to solve the problem, the solution was as follows

Suppose there is an observer on the wedge and the wedge is accelerating towards the left with acceleration $a$. Then the psuedo force on the block of mass $m$ is $ma$.

Balancing forces on $m$ normal to the plane we get $$N + ma\sin\theta = mg\cos\theta$$ Balancing forces on $M$ in $X$ direction $$F + N\sin\theta = F\cos\theta + Ma$$

Solving these two equations gives me the solution to the above question as follows

$$a = \frac{mg\sin\theta\cos\theta + F(1-\cos\theta)}{M+m\sin^2\theta}$$

This does match with my answer and with the answer in my textbook, however i do not understand why there would be a psuedo force on the wedge of mass $M$, amounting to $Ma$.

With respect to a observer on the wedge should there be no psuedo force on it?.

Any help would be highly appreciated!

Pseudo force as seen by an accelerating frame of refrence

A question struck my mind when i was trying to solve the following problem, 

Problem statement

I was able to solve it by just considering forces in the horizontal and vertical direction however the solution turned out to be very lengthy, i found a solution for it online which used psuedo/fictious forces to solve the problem, the solution was as follows

Suppose there is an observer on the wedge and the wedge is accelerating towards the left with acceleration $a$. Then the psuedo force on the block of mass $m$ is $ma$.

Balancing forces on $m$ normal to the plane we get $$N + ma\sin\theta = mg\cos\theta$$ Balancing forces on $M$ in $X$ direction $$F + N\sin\theta = F\cos\theta + Ma$$

Solving these two equations gives me the solution to the above question as follows

$$a = \frac{mg\sin\theta\cos\theta + F(1-\cos\theta)}{M+m\sin^2\theta}$$

This does match with my answer and with the answer in my textbook, however i do not understand why there would be a psuedo force on the wedge of mass $M$, amounting to $Ma$.

With respect to a observer on the wedge should there be no psuedo force on it?.

Any help would be highly appreciated!

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Pratham Pawan
Pratham Pawan
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Psuedo force as seen by a accelerating frame of refrence

A question struck my mind when i was trying to solve the following problem, Problem statement

I was able to solve it by just considering forces in the horizontal and vertical direction however the solution turned out to be very lengthy, i found a solution for it online which used psuedo/fictious forces to solve the problem, the solution was as follows

Suppose there is an observer on the wedge and the wedge is accelerating towards the left with acceleration $a$. Then the psuedo force on the block of mass $m$ is $ma$.

Balancing forces on $m$ normal to the plane we get $$N + ma\sin\theta = mg\cos\theta$$ Balancing forces on $M$ in $X$ direction $$F + N\sin\theta = F\cos\theta + Ma$$

Solving these two equations gives me the solution to the above question as follows

$$a = \frac{mg\sin\theta\cos\theta + F(1-\cos\theta)}{M+m\sin^2\theta}$$

This does match with my answer and with the answer in my textbook, however i do not understand why there would be a psuedo force on the wedge of mass $M$, amounting to $Ma$.

With respect to a observer on the wedge should there be no psuedo force on it?.

Any help would be highly appreciated!