Question(9) A body weighing W newton is placed on a smooth plane inclined at to the horizontal kept in equilibrium under the action of a horizontal force of magnitude 33 newton , then the weight of the body W = ........... newton. (a) 38 (b) 33 (c) 46.66 (d) 75.15
Studdy Solution
STEP 1
What is this asking? Find the weight of a body on a 45° inclined plane, balanced by a 33 N horizontal force. Watch out! Don't forget to consider the components of the forces along the plane!
STEP 2
1. Break down the forces
2. Set up the equilibrium equation
3. Solve for
STEP 3
First, let's **identify the forces** acting on the body.
We have the weight acting vertically downwards and a horizontal force of **33 N**.
STEP 4
Since the plane is inclined at **45°**, we need to **resolve the weight** into two components: one parallel to the plane and one perpendicular to the plane.
The parallel component is and the perpendicular component is .
STEP 5
The horizontal force also needs to be resolved into components along the plane.
The component of the horizontal force along the plane is .
STEP 6
For the body to be in **equilibrium**, the net force along the plane must be zero.
So, the component of the weight along the plane should equal the component of the horizontal force along the plane.
STEP 7
Set up the equation: \[ W \cdot \sin(45^\circ) = 33 \cdot \cos(45^\circ) $
STEP 8
Since , we can simplify the equation to: \[ W \cdot \frac{\sqrt{2}}{2} = 33 \cdot \frac{\sqrt{2}}{2} $
STEP 9
Divide both sides by to solve for : \[ W = 33 $
STEP 10
The weight of the body is **33 newtons**.
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