A box weighing 5kg sits on a box weighing 10kg. The static friction between them is .5. There is no friction under the bottom box. How much tension is needed on the bottom box to make the top box slide?
I assume you mean the static coefficent of friction is .5
Given that, then the force of friction static on the top box is .5(5)g max. over that, and it slides.
.5(5g)<ma
.5*5g<5a
or a=.5g
Now to achieve that acceleration, then
tension=totalmass*.5g=(15)(.5g)
To determine the amount of tension required to make the top box slide off the bottom box, we need to consider the forces acting on the system.
First, we need to calculate the maximum static friction force between the boxes. The static friction force can be given by the formula:
Fs ≤ μs * N
where Fs is the static friction force, μs is the coefficient of static friction, and N is the normal force.
In this case, the normal force acting on the top box is its weight, which is given by:
N = m * g
where m is the mass of the top box and g is the acceleration due to gravity (approximately 9.8 m/s^2).
Given that the mass of the top box is 5 kg, the normal force acting on it is:
N = 5 kg * 9.8 m/s^2 = 49 N
Now, we can calculate the maximum static friction force:
Fs ≤ μs * N
Fs ≤ 0.5 * 49 N
Fs ≤ 24.5 N
Therefore, the maximum static friction force between the boxes is 24.5 N. This means that a tension force greater than 24.5 N is required to overcome the static friction and make the top box slide.
In this case, the tension force acting on the bottom box will be the same as the force required to overcome the static friction force. Therefore, a tension force of greater than 24.5 N is needed on the bottom box to make the top box slide.