A block of mass
M1 = 2.7 kg
rests on top of a second block of mass
M2 = 4.7 kg,
and the second block sits on a surface that is so slippery that the friction can be assumed to be zero (see the figure below).
(a) If the coefficient of static friction between the blocks is
¦ÌS = 0.21,
how much force can be applied to the top block without the blocks slipping apart?
Incorrect: Your answer is incorrect.
N
(b) How much force can be applied to the bottom block for the same result?
(a)
Top block:
F-friction=M1a
F=2.7a+(0.21*2.7*9.8)
From bottom block:
friction=M2a
a=(0.21*2.7*9.8)/4.7
F=(0.21*2.7*9.8*2.7)/4.7+(0.21*2.7*9.8)=8.749N
(b) Bottom block:
F-friction=M2a
F=4.7a+(0.21*2.7*9.8)
Top block: 0.21*2.7*9.8=2.7a
a=0.21*9.8
F=(0.21*4.7*9.8)+(0.21*2.7*9.8)=15.23N
To find the force that can be applied to the top block without the blocks slipping apart, we can use the concept of equilibrium. When the blocks are in equilibrium, the force applied to the top block must be equal to the maximum static friction force between the two blocks.
Let's start by finding the maximum static friction force between the blocks. The formula for static friction is:
Fstatic = μS * N
Where:
- Fstatic is the maximum static friction force
- μS is the coefficient of static friction
- N is the normal force between the blocks
In this case, the normal force N can be found by considering the forces acting on the lower block. The weight of the lower block (M2) will be balanced by the normal force and force applied on it. The weight is given by:
Weight = M2 * g
Where:
- M2 is the mass of the lower block
- g is the acceleration due to gravity
Since the surface is slippery and no friction acts on the bottom block, the normal force (N) is equal to the weight of the lower block.
Now let's calculate the normal force between the blocks:
N = Weight = M2 * g
Substituting the given values, we have:
M2 = 4.7 kg (mass of the lower block)
g = 9.8 m/s² (acceleration due to gravity)
N = 4.7 kg * 9.8 m/s² = 46.06 N
Now we can find the maximum static friction force:
Fstatic = μS * N
Substituting the given coefficient of static friction:
μS = 0.21 (coefficient of static friction)
Fstatic = 0.21 * 46.06 N
Fstatic = 9.653 N
So, the maximum force that can be applied to the top block without the blocks slipping apart is approximately 9.653 N.
Now let's move on to part (b), which asks for the force that can be applied to the bottom block for the same result.
Since the blocks are in equilibrium, the force applied to the top block is equal to the force applied to the bottom block. Therefore, the maximum force that can be applied to the bottom block without the blocks slipping apart is also 9.653 N, as calculated above.