Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Get precise and detailed answers to your questions from a knowledgeable community of experts on our Q&A platform.
Sagot :
Answer:
a) 113N
b) 0.37
Explanation:
a) Using the Newton's second law:
\sum Fx =ma
Since the crate is not moving then its acceleration will be zero. The equation will become:
\sum Fx = 0
\sumFx = 0
Fm - Ff = 0.
Fm is the moving force
Ff is the frictional force
Fm = Ff
This means that the moving force is equal to the force of friction if the crate is static.
Since applied force is 113N, hence the magnitude of the static friction force will also be 113N
b) Using the formula
Ff = nR
n is the coefficient of friction
R is the reaction = mg
m is the mass of the crate = 31.2kg
g is the acceleration due to gravity = 9.8m/s²
R = 31.2 × 9.8
R = 305.76N
Recall that;
n = Ff/R
n = 113/305.76
n = 0.37
Hence the minimum possible value of the coefficient of static friction between the crate and the floor is 0.37
Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Thank you for choosing our platform. We're dedicated to providing the best answers for all your questions. Visit us again. Westonci.ca is here to provide the answers you seek. Return often for more expert solutions.