A block of mass 20kg rests on a smooth horizontal ground near to a fixed post, to which it is attached to a light, horizontal rod. The block is pulled away from the post by force PN inclined at 30 degrees to the horizontal.
Find in terms of P, the tension of the rod and the normal reaction between the block and the ground.
Explain what would happen if the value of p exceeded 392.
I know that b is something like the block would lift of the ground but do not know how to do part a
the horizontal component of P (along the rod) is [P * cos(30º)]
the vertical component is [P * sin(30º)]
the normal reaction is the block's weight (m * g), minus the vertical component of P
you surmise correctly re: the result of P exceeding 392 Newtons
To find the tension in the rod and the normal reaction between the block and the ground, we need to analyze the forces acting on the block.
Here's a step-by-step explanation of how to calculate these values:
Step 1: Resolve the force PN into its horizontal and vertical components.
The horizontal component, PH, is given by PH = PN * cos(30°).
The vertical component, PV, is given by PV = PN * sin(30°).
Step 2: Determine the horizontal forces acting on the block.
Since the block rests on a smooth horizontal surface, there is no friction. Therefore, the only horizontal force acting on the block is the tension in the rod, which pulls the block towards the post. Hence, the tension in the rod is equal to the horizontal component of the force PN. Therefore, the magnitude of the tension in the rod is equal to the magnitude of PH, which is PH = PN * cos(30°).
Step 3: Find the normal reaction between the block and the ground.
The normal reaction is the perpendicular force exerted by the ground on the block to support its weight. Since the block is not moving vertically, the normal reaction force must balance the vertical forces acting on the block.
The total vertical forces acting on the block are the weight of the block (mg) and the vertical component of the force PN (PV).
Therefore, the magnitude of the normal reaction force is equal to the sum of the magnitudes of the weight and PV. Hence, the magnitude of the normal reaction is RN = mg + PV = 20 * 9.8 + PN * sin(30°).
Now, let's move on to explaining what would happen if the value of P exceeded 392.
If the value of P exceeds 392, the tension in the rod would exceed the weight of the block. This means that the upward force provided by the tension in the rod would be greater than the downward force due to the weight of the block. As a result, the block would start to lift off the ground.
In this situation, the normal reaction between the block and the ground would decrease and eventually become zero when the block completely lifts off the ground. The tension in the rod would continue to increase until it reaches a maximum value, after which it cannot increase any further as it would exceed the force applied by P.
It's important to note that this analysis assumes an ideal situation with no external factors or restrictions impacting the movement of the block. In reality, factors such as the strength of the fixed post and the rigidity of the rod could also play a role and may impose limitations on the system.