3. A museum curator moves artifacts into place on various different display surfaces.
Use the values in Figure 4.7 to find Fs,max and F, for the following situations:
a. moving a 145 kg aluminum sculpture across a horizontal steel platform
b. pulling a 15 kg steel sword across a horizontal steel shield
c. pushing a 250 kg wood bed on a horizontal wood floor
d. sliding a 0.55 kg glass amulet on a horizontal glass display case
To find Fs,max and F for the given situations, we will use the following formula:
Fs,max = μs * N
F = μk * N
a. Moving a 145 kg aluminum sculpture across a horizontal steel platform:
Given:
Mass of the sculpture, m = 145 kg
Coefficient of static friction, μs = 0.6
Coefficient of kinetic friction, μk = 0.4
To find Fs,max:
First, we need to find the normal force, N, acting on the sculpture.
The weight of the sculpture, W = m * g, where g is the acceleration due to gravity.
Assuming g ≈ 9.8 m/s^2, we have W = 145 kg * 9.8 m/s^2 = 1421 N.
Since the sculpture is on a horizontal surface, the normal force is equal to the weight, N = 1421 N.
Now, we can calculate Fs,max:
Fs,max = μs * N = 0.6 * 1421 N ≈ 852.6 N
To find F:
F = μk * N = 0.4 * 1421 N ≈ 568.4 N
b. Pulling a 15 kg steel sword across a horizontal steel shield:
Given:
Mass of the sword, m = 15 kg
Coefficient of static friction, μs = 0.75
Coefficient of kinetic friction, μk = 0.6
To find Fs,max:
The normal force, N, acting on the sword is equal to its weight, N = m * g = 15 kg * 9.8 m/s^2 = 147 N.
Now, we can calculate Fs,max:
Fs,max = μs * N = 0.75 * 147 N ≈ 110.25 N
To find F:
F = μk * N = 0.6 * 147 N ≈ 88.2 N
c. Pushing a 250 kg wood bed on a horizontal wood floor:
Given:
Mass of the bed, m = 250 kg
Coefficient of static friction, μs = 0.6
Coefficient of kinetic friction, μk = 0.4
To find Fs,max:
The normal force, N, acting on the bed is equal to its weight, N = m * g = 250 kg * 9.8 m/s^2 = 2450 N.
Now, we can calculate Fs,max:
Fs,max = μs * N = 0.6 * 2450 N = 1470 N
To find F:
F = μk * N = 0.4 * 2450 N = 980 N
d. Sliding a 0.55 kg glass amulet on a horizontal glass display case:
Given:
Mass of the amulet, m = 0.55 kg
Coefficient of static friction, μs = 0.2
Coefficient of kinetic friction, μk = 0.15
To find Fs,max:
The normal force, N, acting on the amulet is equal to its weight, N = m * g = 0.55 kg * 9.8 m/s^2 = 5.39 N.
Now, we can calculate Fs,max:
Fs,max = μs * N = 0.2 * 5.39 N = 1.078 N
To find F:
F = μk * N = 0.15 * 5.39 N = 0.809 N