A person has a choice while trying to move a crate across a horizontal pad of concrete: push it at a downward angle of 30 degrees, or pull it at an upward angle of 30 degrees.
a) Which choice is most likely to require less force on the part of the person?
1. pushing at a downward angle
2. pulling at the same angle, but upward
3. pushing or pulling shouldn't matter
The answer I chose was 2 because this would decrease friction resulting in decrease of normal force.
b) If the crate has a mass of 50.0 kg and the coefficient of friction between it and the concrete is 0.750, calculate the force required to move it across the concrete at a constant speed in both situations.
This is what I don't really get. I was thinking of using Fpcos(30) = uN with N being (mg + Fpsin(30)) for pushing and then Fpcos(30) = uN with N being (mg - Fpsin(30)) for pulling. Any ideas?
Block A, with a mass of 50 kg, rests on a horizontal table top. The coefficient of static friction is 0.40. A horizontal string is attached to A and passes over a massless, ideal pulley as shown.
What is the smallest mass of Block B that will start A moving when it is attached to the other end of the string?
3 answers
Block B pulls on A
so at the point when weight of B equals the friction on A..
massB*g=massA*mu*g
so at the point when weight of B equals the friction on A..
massB*g=massA*mu*g
On the crate, your approach is correct.
upward angle
Fcos30=.75(50g-Fsin30)
f(cos30+.75sin30)=.75*50g
solve for F, and for the downward angle
f(cos30-.75sin30)=.75*50g
upward angle
Fcos30=.75(50g-Fsin30)
f(cos30+.75sin30)=.75*50g
solve for F, and for the downward angle
f(cos30-.75sin30)=.75*50g