Question

Newton's second law: Rudolph the red nosed reindeer is pulling a 25 kg sled across the snow in a field. The coefficient of kinetic friction is .12 The rope that is pulling the sled is coming off at a 29 degree angle above the horizontal. Find the force in the rope when the acceleration is .12 m/s^2.

Answer 1

T= 34.7 N :Force in the rope

Step-by-step explanation:

Look at the sled free body diagram in the attached graphic

T:Force in the rope (N)

Ff= Friction Force (N)

W : Sled weight , W=m*g (N)

N : normal force (N)

m: Sled mass (kg) = 25 kg

g : acceleration due to gravity= 9,8 m/s²

a : acceleration = 0.12 m/s²

μ: coefficient of kinetic friction = 0.12

∑Fx=m*a Newton's second law

Tx-Ff = m*a

Tcos29°-μN=m*a Equation (1)

∑Fy= 0 Newton's first law

N+Ty -W=0

N+Tsin29°-W=0

N=-Tsin29°+W

N= -Tsin29°+m*g

We replace N= -Tsin29°+ m*g in the equation (1)

Tcos29°-μ( -Tsin29°+m*g)=m*a

Tcos29°+μ*Tsin29°´-μm*g=m*a

T(cos29°+0.12*sin29°)= m*a+μm*g=25*0.12+25*0.12*9.8

T(0.9327)=32.4

`T=(32.4)/(0.9327)`

T= 34.7 N