Two perpendicular forces, one of 60N directed upward and one of 70N directed to the right, act simultaneously on an object having a mass of 5kg. The coefficient of friction between the object and...

Two perpendicular forces, one of 60N directed upward and one of 70N directed to the right, act simultaneously on an object having a mass of 5kg. The coefficient of friction between the object and the floor is 0.37, so what is the objects rate of acceleration?

2 Answers | Add Yours

valentin68's profile pic

valentin68 | College Teacher | (Level 3) Associate Educator

Posted on

On the vertical axis the force is

`Fy = 60 -m*g =60-5*9.81 = 10.95 N`

Since the force on the vertical axis is positive it means the object will lift from the very first moment no matter the value of the forces on the horizontal axis. It means there will be no friction force with the ground. Therefore on the horizontal axis

`Fx = 70 N`

Total applied force is the sum of Fxand Fy taken as vectors.

F= Fx+Fy

In absolute value

`F = sqrt(Fx^2 +Fy^2) = sqrt(70^2+10.95^2) = 70.85 N`

The corresponding acceleration is

`a = F/m =70.85/5 = 14.17 m/s^2`

The angle `alpha` that acceleration a is making with the horizontal is

`alpha = arctan((Fy)/(Fx)) =arctan(10.95/70) = 8.89 degree `

counterclockwise from the positive direction of x axis.

maahi1704's profile pic

maahi1704 | Student, Grade 11 | (Level 3) eNoter

Posted on

F-f = ma

[(60*60 + 70*70 )^1/2] - coefficient of friction*m*g = ma

92.19 - 18.13 = 5*a

a=14.8m/s^2

We’ve answered 318,982 questions. We can answer yours, too.

Ask a question