a cart is loaded with a brick and pulled up at a constant speed along an inclined plane to the height of a seattop. if the mass
2 answers:
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Answer:
Explanation:
g = Acceleration due to gravity =
= Angle of slope =
v = Velocity of child at the bottom of the slide
= Coefficient of kinetic friction
= Coefficient of static friction
h = Height of slope = 1.8 m
The energy balance of the system is given by
The speed of the child at the bottom of the slide is
Length of the slide is given by
The force energy balance of the system is given by
The coefficient of kinetic friction is .
For static friction
So, the minimum possible value for the coefficient of static friction is .
Answer:
94.67 N
Explanation:
Consider a free body diagram with force, F of 41 N applied at an angle of 37 degrees while the weight acts downwards. Resolving the force into vertical and horizontal components, we obtain a free body diagram attached.
At equilibrium, normal reaction is equal to the sum of the weight and the vertical component of the force applied. Applying the condition of equilibrium along the vertical direction.
Substituting 70 N for W, 41 N for F and for 37 degrees
N=70+41sin37=94.67441595 N and rounding off to 2 decimal places
N=94.67 N
Answer:
R= 602 .11 N
Explanation:
The horizontal component of tension T will give reaction of the wall and the vertical component of T will balance the weight of of the climber .
T cos32 = R
710 x .848 = R
R= 602 .11 N .
