A block of mass m1 = 4.50 kg sits on top of a second block of mass m2 = 13.9 kg, which in turn is on a horizontal table. The coefficients of friction between the two blocks are µs = 0.300 and µk = 0.100. The coefficients of friction between the lower block and the rough table are µs = 0.500 and µk = 0.400. You apply a constant horizontal force to the lower block, just large enough to make this block start sliding out from between the upper block and the table.
Determine the magnitude of each force on each block at the instant when you have started pushing but motion has not yet started. In particular, what force must you apply?
I tried finding the applied force using:
Fa = µs (m2g) + µs (m1g)
Fa = .5(13.9 x 9.8) + .3(4.5 x 9.8)= 81.34 N. This was not correct. Any suggestions?
To find the force you must apply to start the motion but before it starts, you need to consider the maximum static friction between the lower block and the upper block, as well as between the lower block and the table.
Let's calculate the maximum static friction between the lower block and the upper block:
F_max_static_friction = µs * Normal Force
Normal Force = m1 * g (since both blocks are on the same horizontal table)
F_max_static_friction = µs * m1 * g
F_max_static_friction = 0.300 * 4.50 kg * 9.8 m/s^2
Now, let's calculate the maximum static friction between the lower block and the table:
F_max_friction_table = µs * Normal Force_table
Normal Force_table = m2 * g (since only the lower block is in contact with the table)
F_max_friction_table = µs * m2 * g
F_max_friction_table = 0.500 * 13.9 kg * 9.8 m/s^2
To find the force you must apply, it needs to overcome both of these maximum static frictions. Therefore,
Applied force = F_max_static_friction + F_max_friction_table
Applied force = (0.300 * 4.50 kg * 9.8 m/s^2) + (0.500 * 13.9 kg * 9.8 m/s^2)
Now you can calculate the applied force using the given values and solve for the correct answer.
To find the magnitude of the force (Fa) that needs to be applied to the lower block in order to make it start sliding out from between the upper block and the table, you need to consider the forces acting on both blocks.
Let's break down the problem and analyze the forces acting on each block separately:
For the upper block:
1. Weight force (mg) acts downwards.
2. Normal force (N1) acts upwards, exerted by the lower block.
For the lower block:
1. Weight force (mg) acts downwards.
2. Normal force (N2) acts upwards, exerted by the table.
3. Friction force (f2) acts horizontally in the opposite direction of the applied force.
Now, let's determine the magnitudes of these forces:
For the upper block:
1. Weight force (m1 * g)
2. Normal force (N1 = m1 * g) since the upper block is not accelerating vertically.
For the lower block:
1. Weight force (m2 * g)
2. Normal force (N2 = m2 * g) since the lower block is not accelerating vertically.
3. Friction force (f2 = µs2 * N2) where µs2 is the coefficient of static friction between the lower block and the table.
Since both blocks are not moving vertically, the normal forces (N1 and N2) are equal in magnitude and cancel each other out.
Now, let's calculate the friction force (f2) on the lower block:
f2 = µs2 * N2 = µs2 * (m2 * g)
The applied force (Fa) is equal in magnitude but opposite in direction to the friction force (f2), so we have:
Fa = f2 = µs2 * (m2 * g)
Plugging in the given values:
µs2 = 0.500
m2 = 13.9 kg
g = 9.8 m/s²
Fa = 0.500 * (13.9 * 9.8)
Calculating this, we find:
Fa ≈ 68.17 N
Therefore, the magnitude of the force that must be applied to the lower block in order to make it start sliding is approximately 68.17 N.
Note: It's worth noting that the friction coefficient used for the upper block does not come into play since it is not moving vertically and is kept in place by the normal force exerted by the lower block.