Mass 1 (10 kg) rests on a table connected by a string to Mass 2 (5 kg). Find the minimum coefficient of static friction for which the blocks remain stationary.
I personally do not have much inclination to work your homework for you. I will be happy to critique your work or thinking.
well that was not the point...my teacher just did one problem with us and gave us these review problems...and i searched everywhere in order to figure out how to do these problems...it would be great if you can help at least get started on them and then from there i think i can handle it...it is just the part of getting started since i do not even have any formulas
To find the minimum coefficient of static friction for which the blocks remain stationary, we need to consider the forces acting on each block.
Let's label Mass 1 as M1 (10 kg) and Mass 2 as M2 (5 kg). The tension in the string connecting them will be the same for both blocks, denoted as T.
Since the blocks are stationary, the forces in the vertical direction must balance out. We have:
F_gravity_M1 + F_tension = F_normal_M1
F_gravity_M2 = F_normal_M2
The forces in the horizontal direction must also balance out, otherwise there would be a net force moving the blocks.
The force of friction (F_friction) exerted on Mass 1 due to the table can be calculated as:
F_friction = μ_s * F_normal_M1
Where μ_s is the coefficient of static friction and F_normal_M1 is the normal force acting on Mass 1. Since the blocks are connected by a string, F_friction will also act on Mass 2.
Now let's calculate the forces:
For Mass 1:
F_gravity_M1 = m1 * g
F_normal_M1 = m1 * g
For Mass 2:
F_gravity_M2 = m2 * g
F_normal_M2 = m2 * g
Since the mass of Mass 1 is 10 kg and the mass of Mass 2 is 5 kg, we'll use g = 9.8 m/s^2.
By substituting into the equations, we have:
F_gravity_M1 = (10 kg) * (9.8 m/s^2) = 98 N
F_normal_M1 = (10 kg) * (9.8 m/s^2) = 98 N
F_gravity_M2 = (5 kg) * (9.8 m/s^2) = 49 N
F_normal_M2 = (5 kg) * (9.8 m/s^2) = 49 N
Since the blocks are stationary, the horizontal forces must balance:
F_friction = F_tension
Substituting μ_s * F_normal_M1 for F_friction, we have:
μ_s * F_normal_M1 = F_tension
And substituting μ_s * F_normal_M1 for F_friction on Mass 2, we get:
μ_s * F_normal_M1 = μ_s * F_normal_M2
But since F_normal_M1 = F_normal_M2, canceling out common terms gives:
μ_s = F_normal_M2 / F_normal_M1
Now we can substitute the values:
μ_s = 49 N / 98 N
μ_s = 0.5
Therefore, the minimum coefficient of static friction required for the blocks to remain stationary is 0.5.