A wedge of mass m = 36.1 kg is located on a plane that is inclined by an angle θ = 22.1 with respect to the horizontal. A force F = 308.3 N in horizontal direction pushes on the wedge, as shown. The coefficient of friction between the wedge and the plane is 0.193.
What is the acceleration of m along the plane? Negative numbers for motion to the left, and positive numbers for motion to the right, please.
2007-02-07 10:16:45 · 1 answers · asked by Anonymous in Science & Mathematics ➔ Physics
F=ma= Fh cos(22.1)-uN
ma= Fh cos(22.1)-u mg cos(22.1)
a=(cos(22.1)/m)(Fh-mg)=
a=(cos(22.1)/36.1)(308.3- .193 x36.1 x 9.81)=
a=6.2 m/s
I need the drawing to specify the direction
2007-02-08 05:01:23 · answer #1 · answered by Edward 7 · 0⤊ 1⤋
a) Newton's 2nd Law as it applies to the two mass system. m2*g - m1*g*sin(θ) = (m1 + m2)a Solve for a. b) Newtons's 2nd Law as it applies to m2 alone. T-m2*g=m2*a a is given by the solution to part a) Solve for T=tension.
2016-03-15 08:57:42 · answer #2 · answered by Anonymous · 0⤊ 0⤋