An older gentleman drags a wooden crate (10 kg) out of storage. The 120 N force that he applies on the crate makes an angle of 30 degrees with the horizontal. The coefficient of kinetic friction between the floor and the crate is 0.1. What is the frictional force on the crate?
A. 2.4 N
B. 3.8 N
C. 7.2 N
D. 9.8 N
The answer is B. According to the text, the frictional force is calculated by finding the normal force, which is supposed to be F(normal) = mg - F*sin(30).
I don't understand how this equation was set up. If the upward force is F*sin(30), shouldn't this only add to the F(normal)? So I would think that F*sin(30) = F(normal) - mg.
To me, having an upward force on the crate is sort of like a person in an elevator accelerating upward, where ma = F(normal) - mg.
A. 2.4 N
B. 3.8 N
C. 7.2 N
D. 9.8 N
The answer is B. According to the text, the frictional force is calculated by finding the normal force, which is supposed to be F(normal) = mg - F*sin(30).
I don't understand how this equation was set up. If the upward force is F*sin(30), shouldn't this only add to the F(normal)? So I would think that F*sin(30) = F(normal) - mg.
To me, having an upward force on the crate is sort of like a person in an elevator accelerating upward, where ma = F(normal) - mg.