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In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to?
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In the arrangement shown,all surfaces are frictionless.the rod R is co...
Explanation:
To solve this problem, we need to apply Newton's second law of motion to both the rod and the wedge.
For the rod R:
The only force acting on the rod R is its weight, which is directed downwards. Therefore, the net force acting on the rod is given by:
F_net = m_R * a
where m_R is the mass of the rod and a is its vertical acceleration.
For the wedge W:
The forces acting on the wedge are its weight, which is directed downwards, and the normal force N exerted by the ground on the wedge, which is directed upwards and perpendicular to the ground. The net force acting on the wedge is given by:
F_net = m_W * a
where m_W is the mass of the wedge and a is its horizontal acceleration.
However, we need to relate the acceleration of the rod to that of the wedge. To do this, we can use the fact that the rod is constrained to move vertically. This means that its velocity is always perpendicular to the surface of the wedge. Therefore, the horizontal component of the rod's acceleration must be equal to the horizontal acceleration of the wedge. This gives us:
a_R sin(theta) = a_W
where theta is the angle between the rod and the horizontal.
Now, we can substitute this expression for a_W into the equation for the net force on the wedge:
F_net = m_W * a_W
F_net = m_W * a_R sin(theta)
We can also substitute the expression for F_net from the equation for the net force on the rod:
m_R * a = m_W * a_R sin(theta)
Simplifying this expression, we get:
a/a_R = m_R/m_W sin(theta)
Therefore, the ratio a/a_R is equal to m_R/m_W sin(theta).
To solve this problem, we need to apply Newton's second law of motion to both the rod and the wedge.
For the rod R:
The only force acting on the rod R is its weight, which is directed downwards. Therefore, the net force acting on the rod is given by:
F_net = m_R * a
where m_R is the mass of the rod and a is its vertical acceleration.
For the wedge W:
The forces acting on the wedge are its weight, which is directed downwards, and the normal force N exerted by the ground on the wedge, which is directed upwards and perpendicular to the ground. The net force acting on the wedge is given by:
F_net = m_W * a
where m_W is the mass of the wedge and a is its horizontal acceleration.
However, we need to relate the acceleration of the rod to that of the wedge. To do this, we can use the fact that the rod is constrained to move vertically. This means that its velocity is always perpendicular to the surface of the wedge. Therefore, the horizontal component of the rod's acceleration must be equal to the horizontal acceleration of the wedge. This gives us:
a_R sin(theta) = a_W
where theta is the angle between the rod and the horizontal.
Now, we can substitute this expression for a_W into the equation for the net force on the wedge:
F_net = m_W * a_W
F_net = m_W * a_R sin(theta)
We can also substitute the expression for F_net from the equation for the net force on the rod:
m_R * a = m_W * a_R sin(theta)
Simplifying this expression, we get:
a/a_R = m_R/m_W sin(theta)
Therefore, the ratio a/a_R is equal to m_R/m_W sin(theta).
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In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to?
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In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to?.
In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to? for Class 11 2024 is part of Class 11 preparation. The Question and answers have been prepared according to the Class 11 exam syllabus. Information about In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to? covers all topics & solutions for Class 11 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for In the arrangement shown,all surfaces are frictionless.the rod R is constrained to move vertically.the vertical acceleration of R is a and the horizontal acceleration of the wedge W is a.the ratio a/a is equal to?.
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