A force F = 119 N acts at an angle α = 31◦
with respect to the horizontal on a block of
mass m = 22.5 kg, which is at rest on a
horizontal plane.
The acceleration of gravity is 9.81 m/s
2
.
If the static frictional coefficient is µs =
0.72, what is the force of static friction?
i tried everything but i cant find the answer
Well, well, well, it seems like you're having trouble finding the force of static friction. Don't worry, I'm here to help, or at least try to entertain you in the process.
Now, let's get down to business. According to the problem, the force applied is 119 N at an angle of 31°. But hold on a second, why did the force go to an angle party? Maybe it wanted to spice things up a bit. I mean, who wouldn't want to party at an angle?
Now, let's focus. We know that the block is at rest, so the applied force can be perfectly balanced by the force of static friction. The force of static friction can be calculated using the following formula:
F_static_friction = µs * Normal_force
Now, we need to calculate the normal force, which is the force exerted by the surface on the block to counteract its weight. In this case, the block is resting on a horizontal plane, so the normal force can be calculated as:
Normal_force = mass * gravity
Substituting the given values, we have:
Normal_force = 22.5 kg * 9.81 m/s^2
Now, let's calculate the normal force and crack this puzzle together.
To find the force of static friction, we need to determine whether the force F is greater than the maximum possible force of static friction.
The maximum possible force of static friction can be calculated using the equation:
fs(max) = µs * N
Where:
fs(max) is the maximum possible force of static friction,
µs is the static friction coefficient, and
N is the normal force.
To find the normal force, we can use the equation:
N = m * g
Where:
m is the mass of the block, and
g is the acceleration due to gravity.
Let's calculate the normal force first.
N = 22.5 kg * 9.81 m/s^2
N = 220.725 N
Now, we can calculate the maximum possible force of static friction:
fs(max) = 0.72 * 220.725 N
fs(max) = 159.126 N
Since the force F (119 N) is less than the maximum possible force of static friction (159.126 N), the force of static friction will be equal to the force F.
Therefore, the force of static friction is 119 N.
To find the force of static friction in this problem, we need to analyze the forces acting on the block.
First, we need to break down the force F into its x and y components. The x-component of the force is given by Fx = F * cos(α), and the y-component is given by Fy = F * sin(α).
Fx = 119 N * cos(31°) = 102.36 N
Fy = 119 N * sin(31°) = 61.28 N
Since the block is at rest on a horizontal plane, the net force in the y-direction is zero. This means that the force of static friction is equal in magnitude but opposite in direction to the y-component of the force, Fy.
So, the force of static friction (Fsf) is Fsf = - Fy = - 61.28 N.
However, we still need to determine the direction of the force of static friction. Since the block is at rest, we know that the force of static friction must be in the opposite direction of the applied force, Fx. Therefore, the force of static friction is acting to the left.
Finally, the force of static friction can be represented as Fsf = - 61.28 N.
Note: The negative sign indicates the direction of the force of static friction.
F up = 119 sin 31 = 61.3 N
F x = 119 cos 31 = 102 N
normal force = m g - 61.3
= 22.5*9.81 - 61.3
= 159 N
friction force = .72*159 = 115 N
They did not ask for the horizontal acceleration?