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Question 1
A rough plane is inclined to the horizontal at an angle $ heta$, where $ an heta = rac{3}{4}$. A brick P of mass m is placed on the plane. The coefficient of fri... show full transcript
Step 1
Answer
To find the normal reaction, we first need to resolve the forces acting on brick P. The component of the weight acting perpendicular to the plane is given by:
R = mg imes rac{4}{5}
Therefore, the normal reaction is:
R = mg imes rac{4}{5}
Step 2
Answer
To show that the coefficient of friction is rac{3}{4}, we resolve the forces acting parallel to the plane. The force down the plane due to gravity is:
F = mg imes rac{3}{5}
In equilibrium, the frictional force is equal to the component of the weight down the plane:
u R$$ Substituting for $R$, we have: $$F_{friction} = u imes mg imes rac{4}{5}$$ Setting these equal: $$mg imes rac{3}{5} = u imes mg imes rac{4}{5}$$ Dividing both sides by $mg imes rac{4}{5}$ gives:u = rac{3}{4}$$
Step 3
Answer
The forces acting on brick Q will balance since the mass component cancels out. The friction between brick Q and the plane will increase in direct proportion to the weight component acting down the plane, thereby preventing any motion.
Step 4
Answer
Brick Q slides down the plane with constant speed. This indicates that there is no resultant force acting on it along the plane since the frictional force equals the component of its weight acting down the slope. As a result, there is no acceleration.
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