1. Find Fn
Fn=18N*cos(28)=15.9N
2. Find Forces in Y direction to solve for m
18N*sin(28)+Fn*fs
18N*sin(28)+(15.9N*0.4)-9.8m=0
Ans: m=1.5kg
A small box is held in place against a rough vertical wall by someone pushing on it with a force directed upward at 28degrees above the horizontal. The coefficients of static and kinetic friction between the box and wall are 0.40 and 0.30, respectively. The box slides down unless the applied force has magnitude 18 N.
What is the mass of the box in kilograms?
5 answers
where did you get the 0.4??
That's the static coefficient of friction for the surface
2. Find Forces in Y direction to solve for m
18N*sin(28)+Fn*fs
18N*sin(28)+(15.9N*0.4)-9.8m=0
The above solution method is incorrect. I assume the intended 18N*sin(28)+(15.9N*0.4) / 9.8 = m.
The forces in the y direction divided by the acceleration would net you 1.511274299 approx mass for the object.
18N*sin(28)+Fn*fs
18N*sin(28)+(15.9N*0.4)-9.8m=0
The above solution method is incorrect. I assume the intended 18N*sin(28)+(15.9N*0.4) / 9.8 = m.
The forces in the y direction divided by the acceleration would net you 1.511274299 approx mass for the object.
Where did the 15.9N come from?
0 answers