There are Two boxes on top of each other. The top has a mass of 2.99kg the bottom one has a mass of 7.89 kg. A horizontal force of 39.97 newtons is applied to the left to move them both at an acceleration of 1.42m's^2
Find the minimum coefficient of static friction μt between the blocks such that the 2.99 kg block does not slip under an acceleration of 1.42 m/s2.
Where is the force applied, Top or bottom box?
the force is pulling both boxes to the right
To find the minimum coefficient of static friction (μt) between the blocks such that the 2.99 kg block does not slip under an acceleration of 1.42 m/s^2, we need to consider the forces acting on the blocks.
Let's analyze the forces acting on the system:
1. The applied force (Fa) with a magnitude of 39.97 Newtons is acting horizontally to the left.
2. The force of static friction (Fs) is acting between the two blocks, opposing the applied force.
3. The gravitational force (Fg) is acting vertically downwards on both blocks.
First, let's calculate the net force acting on the system:
Net force (Fnet) = Fa - Fs
The net force is responsible for the acceleration of the system. We can calculate the net force by using Newton's second law of motion:
Fnet = m * a
where m is the total mass of the system and a is the acceleration.
The total mass of the system is the sum of the masses of the two blocks:
Total mass (m) = mass of the top block + mass of the bottom block
m = 2.99 kg + 7.89 kg
Now, substitute the values into the equation for Fnet:
Fnet = (2.99 kg + 7.89 kg) * 1.42 m/s^2
Next, rearrange the equation to solve for Fs:
Fs = Fa - Fnet
Fs = 39.97 N - [(2.99 kg + 7.89 kg) * 1.42 m/s^2]
Now we know the force of static friction (Fs). The force of static friction can be expressed as the coefficient of static friction (μt) multiplied by the normal force (Fn):
Fs = μt * Fn
In this case, the normal force (Fn) is equal to the weight of the top block (2.99 kg) plus the weight of the bottom block (7.89 kg) due to the contact between them:
Fn = (2.99 kg + 7.89 kg) * g
where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Finally, substitute the values into the equation for Fs:
39.97 N - [(2.99 kg + 7.89 kg) * 1.42 m/s^2] = μt * [(2.99 kg + 7.89 kg) * g]
Now, rearrange the equation to solve for μt:
μt = [39.97 N - (10.88 kg * 1.42 m/s^2)] / [(10.88 kg) * 9.8 m/s^2]
By substituting the given values, you can calculate the minimum coefficient of static friction (μt) between the blocks such that the 2.99 kg block does not slip under an acceleration of 1.42 m/s^2.