Fap = 340 N.[20o] Above the hor.
Mass = 54 kg.
u(k) = 0.4
m*g = 54 * 9.8 = 529.2 N. = Wt. of object.
Fn = m*g+Fap*sin20=529.2 N. + 340*Sin20 = 645.5 N. = Normal = Force perpendicular to the surface.
Fk = u*Fn = 0.4 * 645.5 = 258.2 N.
a = (Fap*Cos20-Fk)/m = (319.5-258.2)/54=
1.14 m/s.
Could anyone help me set up this equation? I drew my force body diagram but wasn't sure how to include the downward force (im assuming i use either sine or cosine?)
A force of 340 N pushes a 54 kg object. Find the acceleration of the box if the force makes an angle of 70 degrees from the vertical (the force is pushing down on the box). The coefficient of kinetic friction is 0.4.
2 answers
Fn:340(sin20)(54*9.8)=645.5
fk:.4(645.5)=258.2
a:(340(sin20)-258.2)/54=1.14
fk:.4(645.5)=258.2
a:(340(sin20)-258.2)/54=1.14
2 answers