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.
To find the tension in the rod and the normal reaction between the block and the ground, we can start by analyzing the forces acting on the block.
The force of gravity acting vertically downward can be split into two components: one perpendicular to the ground, known as the normal reaction force (N), and the other parallel to the ground, known as the weight of the block (mg).
Since the block is smooth and there is no friction, there is no parallel force acting on the block. Therefore, the only horizontal force acting on the block is the force (PN) applied at an angle (θ = 30 degrees) to the horizontal.
Using Newton's second law of motion, we can analyze the forces in the x and y directions separately:
In the x-direction:
PN × cos(30°) = Tension in the rod
In the y-direction:
PN × sin(30°) = Normal reaction force (N)
mg = Normal reaction force (N)
Since there is no vertical acceleration, the weight mg and normal reaction force N are equal in magnitude.
Now, let's solve for the tension in the rod and the normal reaction force in terms of P:
From the x-direction:
PN × cos(30°) = Tension in the rod
From the y-direction:
PN × sin(30°) = Normal reaction force (N) = mg
Since the mass (m) is given as 20 kg, the weight (mg) is:
mg = 20 kg × 9.8 m/s^2 = 196 N
Substituting the known values into the equations:
PN × cos(30°) = Tension in the rod
PN × sin(30°) = 196 N
Dividing the second equation by the first equation will give us the ratio between the normal reaction (N) and the tension in the rod:
sin(30°) / cos(30°) = N / Tension in the rod
Simplifying:
tan(30°) = N / Tension in the rod
Since tan(30°) is equal to 1/√3, we can substitute this value:
1/√3 = N / Tension in the rod
Now, we can isolate the tension in the rod (T):
Tension in the rod = N / (1/√3) = √3 × N
Therefore, the tension in the rod is √3 times the normal reaction force (T = √3 × N), and the normal reaction force is equal to the weight of the block (N = mg = 196 N).
If the value of P exceeds 392, the block will start to move away from the fixed post. This is because the force applied (PN) would overcome the friction or any other resistive force that holds the block in its place. Once the block starts moving, the normal reaction force will decrease, and the tension in the rod will no longer be equal to √3 times the normal reaction force.
In this case, if P exceeds 392, the block will not only be pulled away from the post but also move independently without any constraint from the rod.
To find the tension in the rod and the normal reaction between the block and the ground, we can start by drawing a free body diagram of the block.
Let T be the tension in the rod and N be the normal reaction between the block and the ground.
From the free body diagram, we can see that the gravitational force acting on the block is balanced by the vertical component of the tension and the normal reaction.
The gravitational force acting on the block (mg) is given by:
mg = (20 kg) * (9.8 m/s^2) = 196 N
The vertical component of the tension is T sin(30°).
Therefore, we can write the equation for the vertical direction:
N + T sin(30°) = mg
Substituting the value of mg, we get:
N + T sin(30°) = 196 N
Now, to find the horizontal component of the tension, we use T cos(30°).
Therefore, the equation for the horizontal direction is:
T cos(30°) = P
Now we have two equations:
N + T sin(30°) = 196 N ----(1)
T cos(30°) = P ----(2)
To solve these equations, we can multiply equation (2) by sin(30°) and subtract it from equation (1). This will eliminate the T term.
N + T sin(30°) - T sin(30°) sin(30°) = 196 N - P sin(30°)
Simplifying:
N - T sin^2(30°) = 196 N - P * (1/2)
N - T * (1/4) = 196 N - P/2
Multiplying both sides by 4 to eliminate the fractional term:
4N - T = 784 N - 2P
So, we have:
T = 784 N - 4N + 2P
Simplifying further, we get:
T = 780 N + 2P
Therefore, the tension in the rod is given by 780 N + 2P.
To find the normal reaction N, we can substitute the value of T from equation (2) into equation (1):
N + (P / cos(30°)) * sin(30°) = 196 N
N + (P / √3) * (1/2) = 196 N
N + P / (2√3) = 196 N
Subtracting P / (2√3) from both sides:
N = 196 N - P / (2√3)
Multiplying both sides by √3 to eliminate the fractional term:
√3N = 3 * 196 N - P / 2
√3N = 588 N - P / 2
Therefore, the normal reaction N is given by 588 N - P / 2.
Now, let's address what would happen if the value of P exceeded 392.
In our calculations above, we used P to represent the horizontal component of the tension. If P exceeds 392 N, it means that the force PN becomes larger than the maximum static friction force between the block and the ground.
When the force of PN exceeds the maximum static friction force, the block will start to move and will continue to move with a constant velocity, as there is no longer a force opposing its motion. In this case, the normal reaction N between the block and the ground will be equal to the weight of the block (mg).
So if P exceeds 392, the block will start to move and the normal reaction N will be equal to 196 N.