Macie pulls a 40kg rolling truck by a strap angled at 30 degrees above the horiz

Question

Macie pulls a 40kg rolling truck by a strap angled at 30 degrees above the horizontal. The truck is accelerating at 0.15m/s^2. If mu sub r= 0.2 what is the magnitude of the normal force acting on the truck and what is the tension on the strap?

I'm getting stuck on what formula I should use for which part I'm thinking that we have to use equivalency statements to get down to one variable "T" and then use "T" to solve for "Fr" and then finally "Fr" and mu that we're given to solve for the normal force but I'm not really sure that I worked this problem correctly and I'm pretty sure my units are wrong.

Explanation / Answer

Solution) given

Mass m=40 kg

Angle (theeta)=30°

Acceleration a=0.15m/s^2

my sub r (Ur)=0.2

Normal force N=?

Tension T=?

We have

Sigma Fx=0

Tcos(theeta) - fr=ma

Sigma Fy=0

Tsin(theeta) - mg=0

Tsin(theeta)=mg

Tsin(30)=40×9.8

T=784 N

Similarly

Tcos(theeta) - fr=ma

784cos(30) - fr =40×0.15

784cos(30) - 40×0.15 =fr

672.96=fr

fr=Ur(N)

N is normal force

N=fr/Ur

N=672.96/0.2=3364.8 N

Normal force N =3364.8N

Tension T=784N

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