Answer: F = 131.7N
Explanation:
You are given the following parameters.
Mass M = 30 kg
Coefficient of static friction μ = 0.5
Ø = 30 degrees
When the person is trying to drag the box with force F, the static frictional force Fs will be acting in the opposite direction.
From the figure attached, resolve all forces into horizontal component and vertical component.
Horizontal component:
Fs - F cosØ = 0
Fs = F cosØ
F cosØ = μN ...... (1)
Vertical component:
N + F SinØ - mg = 0
N = Mg - F SinØ ..... (2)
Substitutes m, g and Ø into the equation 2
N = (30 × 9.8) - F × sin30
N = 294 - 0.5F
Substitute N and coefficient of friction into the equation (1)
F cos30 = 0.5 (294 - 0.5F)
Open the bracket
0.8660F = 147 - 0.25F
Collect the like terms
1.116025F = 147
F = 147/1.116025
F = 131.7 N
Therefore, the minimum force the person needs to have to move the box along the floor is 131.7 N
Answer:
W = 290.7 dynes*cm
Explanation:
d = 1/5 cm = 0.2 cm
The force is in function of the depth x:
F(x) = 1000 * (1 + 2*x)^2
We can expand that as:
F(x) = 1000 * (1 + 4*x + 4x^2)
F(x) = 1000 + 4000*x + 4000*x^2
Work is defined as
W = F * d
Since we have non constant force we integrate
W = [1000*x + 2000*x^2 + 1333*X^3] evaluated between 0 and 0.2
W = 1000*0.2 + 2000*0.2^2 + 1333*0.2^3 - 1000*0 - 2000*0^2 - 1333*0^3
W = 200 + 80 + 10.7 = 290.7 dynes*cm