Question

A 1190 kg car pulls a 460 kg trailer. The car exerts a horizontal force of 4070 N against the ground in order to accelerate. What force does the car exert on the trailer? Assume an effective friction coefficient of 0.190 for the trailer.

Answer 1

F= 4070 N ( the force of the car )
m = 1190 kg
M = 460 kg
k ( or usually the Greek letter:"mi")=0.190
T = ? ( the force on the trailer )
F = ( M +m ) a + k m g
T = m a + k m g
a = ?
a = (F - k m g)/ (M + m )
T = m · (F - k m g ) / ( M + m ) + k m g
T = 1190 kg·( 4070 N-0.190·1190 kg·9.81 m/s²):(460kg+1190kg)+
+0190·1190kg·9.81m/s²=
= 1190 kg ·1851.96 N :1650 kg + 2218.04 N =
=1335.66 N + 2218.04 N = 3553.7 N
The force that car exerts on the trailer is 3553.7 N.