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:
The frequencies are 13.8 Hz, 75 Hz, 12 Hz and 63.8 Hz.
Explanation:
Given that,
The frequency in r.p.m
Suppose, we find the frequency in hz.
We know that,
One r.p.m is equal to the one divided by 60 Hz.
We need to calculate the frequency in Hz
Using formula for frequency in Hz
For f₁,
For f₂,
For f₃,
For f₄,
Hence, The frequencies are 13.8 Hz, 75 Hz, 12 Hz and 63.8 Hz.
Answer:
Dry luricants
Despite being in a solid phase they are able to reduce friction
Hi there!
In this instance, the object spinning in a horizontal circle will experience a net force in the horizontal direction due to tension.
The net force is equivalent to the centripetal force, so:
∑F = T
mv²/r = T
Solve for v:
v = √rT/m
v = 13.96 m/s