Horizontal force is needed to pull the sled

[vicervixen]

A 12 kg sled is pulled along at a constant velocity on a horizintal surface by a horizontal force of 11N. How much horizontal force is needed to pull the sled at a constant velocity it two 57 kg girls are sitting in it?
How would I go about solving this?
Thanks, Amanda

 

[myself]

Draw a free body diagram for the first case. Notice that the sled is in equilibrim because it is moving at a constant velocity. Therefore, all forces sum up to zero. Then, it is possible to figure out the coefficient of friction, which is used to solve the problem.

 

[Brian Earley]

Is the answer 116 N?

 

[whozum]

Total mass = 12kg
Force of Gravity = 12x9.8 = 117.6N
Normal force = -Fg = -117.6N

Net vertical Force = Force of Gravity + Normal force = 117.6 + (-117.6) = 0N
Net Horizontal Force = Pull force + Friction force = 11N + (mu_k)*N = 0N since velocity is constant, there is no acceleration.

11 - mu_k*117.6 = 0
-117.6mu_k = -11

mu_k = -11/-117.6 = 0.1

New mass is 12+2x57 = 126kg
Gravity Force = - Normal force = 1234.8N
Fy = 0
Fx = Pull force + Friction Force = 0
Friction force = mu_k * N = 123.4N
Pull force + 123.4 = 0
Pull force = -123.4N.

The negative sign just means its in the opposite direction of the friction force.

*Editted for new mass

 

[Brian Earley]

Total mass = 12kg
Force of Gravity = 12x9.8 = 117.6N
Normal force = -Fg = -117.6N

Net vertical Force = Force of Gravity + Normal force = 117.6 + (-117.6) = 0N
Net Horizontal Force = Pull force + Friction force = 11N + (mu_k)*N = 0N since velocity is constant, there is no acceleration.

11 - mu_k*117.6 = 0
-1136mu_k = -11

mu_k = -11/-117.6 = 0.1

New mass is 12+2x57 = 116kg
Gravity Force = - Normal force = 1136.8N
Fy = 0
Fx = Pull force + Friction Force = 0
Friction force = mu_k * N = 113.7N
Pull force + 113.7 = 0
Pull force = -113.7N.

The negative sign just means its in the opposite direction of the friction force.

New mass is 126kg.