A uniform sign is supported at P as shown in the figure. If the sign is a square 0.8 m on its side and its mass is 10 kg, what is the magnitude of the horizontal force that P experiences?

ANSWER:

25 N
49 N
98 N
0 N

To find the magnitude of the horizontal force at P, we need to calculate the torque on the sign at P and then use Newton's second law.

The torque on an object is given by the equation torque = force x distance. In this case, the distance is the perpendicular distance between the pivot point P and the center of mass of the sign.

For a square sign, the center of mass lies at the midpoint of the diagonal. The diagonal of a square can be found using the Pythagorean theorem: diagonal^2 = side^2 + side^2.

diagonal = √(0.8m)^2 + (0.8m)^2
diagonal = √(0.64m^2 + 0.64m^2)
diagonal = √1.28m^2
diagonal ≈ 1.13m

The perpendicular distance between P and the center of mass is half the diagonal length: 1.13m / 2 = 0.565m.

Now we can calculate the torque:

torque = force x distance
torque = force x 0.565m

According to Newton's second law, torque is equal to the moment of inertia (I) times the angular acceleration (α). For a square sign rotating about its diagonal, the moment of inertia is 1/6 times the mass times the diagonal length squared.

Let's calculate the moment of inertia:

moment of inertia = 1/6 x mass x diagonal^2
moment of inertia = 1/6 x 10kg x (1.13m)^2

Now we can equate torque and moment of inertia:

torque = moment of inertia x α

Since the sign is in equilibrium (not rotating), the sum of the torques about any point must be zero. Therefore, the torque caused by the gravitational force acting at the center of gravity (which is at the center of mass) must be balanced by the torque caused by the horizontal force at P.

The torque caused by the gravitational force can be calculated using the formula torque = force x perpendicular distance. The force of gravity acting on the sign is equal to the weight of the sign, which is given by the equation weight = mass x gravitational acceleration.

Let's calculate the torque caused by the gravitational force:

torque_gravity = weight x perpendicular distance
torque_gravity = (mass x gravitational acceleration) x perpendicular distance

The mass of the sign is given as 10 kg, and the gravitational acceleration is approximately 9.8 m/s^2.

torque_gravity = (10kg x 9.8m/s^2) x 0.565m

Since the sign is in equilibrium, the torque caused by the horizontal force at P must be equal in magnitude and opposite in direction to the torque caused by the gravitational force.

We can now equate the two torques:

torque = torque_gravity

force x 0.565m = (10kg x 9.8m/s^2) x 0.565m

Simplifying, we find:

force ≈ (10kg x 9.8m/s^2)

Calculating this, we get:

force ≈ 98 N

Therefore, the magnitude of the horizontal force that P experiences is 98 N.

Answer: 98 N

To find the magnitude of the horizontal force that point P experiences, we need to consider the torque acting on the sign. The torque is caused by the weight of the sign, which acts at its center of mass.

The torque equation is given by:
τ = r * F * sin(θ)

Where:
τ = torque
r = distance between the point of rotation (P) and the line of force application
F = force applied
θ = angle between the force and the lever arm

In this case, the weight of the sign acts vertically downwards, so the angle θ between the force and the lever arm is 90 degrees (sin(90) = 1). Also, since the sign is in uniform equilibrium, the sum of the torques acting on it is zero.

The torque due to the weight of the sign is given by:
τ = r * Fg * sin(90)

The force of gravity (Fg) on the sign is given by its mass (m) multiplied by acceleration due to gravity (g):
Fg = m * g

Substituting this back into the torque equation, we have:
τ = r * (m * g) * sin(90)

The magnitude of the horizontal force on P is equal to the torque divided by the lever arm:
F = τ / r

Now we can plug in the given values:
- The length of the side of the square sign (r) = 0.8 m
- The mass of the sign (m) = 10 kg
- The acceleration due to gravity (g) = 9.8 m/s^2

So, the torque is:
τ = 0.8 m * (10 kg * 9.8 m/s^2) * sin(90)
= 0.8 m * 98 N * 1
= 78.4 N*m

And the magnitude of the horizontal force is:
F = 78.4 N*m / 0.8 m
= 98 N

Therefore, the magnitude of the horizontal force that P experiences is 98 N. So the correct answer is:
98 N

49