Answer:
μ = 0.692
Explanation:
In order to solve this problem, we must make a free body diagram and include the respective forces acting on the body. Similarly, deduce the respective equations according to the conditions of the problem and the directions of the forces.
Attached is an image with the respective forces:
A summation of forces on the Y-axis is performed equal to zero, in order to determine the normal force N. this summation is equal to zero since there is no movement on the Y-axis.
Since the body moves at a constant speed, there is no acceleration so the sum of forces on the X-axis must be equal to zero.
The frictional force is defined as the product of the coefficient of friction by the normal force. In this way, we can calculate the coefficient of friction.
The process of solving this problem can be seen in the attached image.
Answer:
no becaus force is mass multiplied by acceleration. the mass of the elephant does not change
Answer:
Explanation:
It is required that the weight of Joe must prevent Simon from being pulled down . That means he is not slipping down but tends to be towed down . So in equilibrium , force of friction will act in upward direction on Simon.
Let in equilibrium , tension in rope be T
For balancing Joe
T = M g
For balancing Simon
friction + T = mgsinθ
μmgcosθ+T = mgsinθ
μmgcosθ+Mg = mgsinθ
M = (msinθ - μmcosθ)
M = m(sinθ - μcosθ)
Answer:
a) the values of the angle α is 45.5°
b) the required magnitude of the vertical force, F is 41 lb
Explanation:
Applying the free equilibrium equation along x-direction
from the diagram
we say
∑Fₓ = 0
Pcosα - 425cos30° = 0
525cosα - 368.06 = 0
cosα = 368.06/525
cosα = 0.701
α = cos⁻¹ (0.701)
α = 45.5°
Also Applying the force equation of motion along y-direction
∑Fₓ = ma
Psinα + F + 425sin30° - 600 = (600/32.2)(1.5)
525sin45.5° + F + 212.5 - 600 = 27.95
374.46 + F + 212.5 - 600 = 27.95
F - 13.04 = 27.95
F = 27.95 + 13.04
F = 40.99 ≈ 41 lb
Answer:
T = 570 N
Explanation:
Given that,
The gravitational force acting on a bucket of water = 525 N
Net force in the Y direction is 45 N
We need to find the magnitude of the force of tension. It can be calculated as :
45 = T - 525
T = 525 + 45
T = 570 N
Hence, the force of tension is 570 N.
