Mass on Bridge, Find Force of Friction when bridge is barely inclined

[nickolaughagus]

Look at the picture attached. There is a mass on a drawbridge, which is inclined at half a meter. We have the Normal force, the coefficient of static friction, force of gravity and the force of friction pushing the mass upwards to the end of the draw bridge that is most inclined. We also have $\theta$, from the joint of the bridge. Put the answer in terms of

What is the force of friction in terms of $\theta$ and force of gravity, F_g
theta and force_gravityThe correct answer is Theta*F_g. How did they get this answer without any trig identities? Did it all cancel out?

 

[PhanthomJay]

It is not sufficient to say that the drawbridge is inclined "half a meter" without giving more info. Perhaps you mean the angle theta is rather small? What do you get for a result? Please show your work.

 

[nickolaughagus]

It is not sufficient to say that the drawbridge is inclined "half a meter" without giving more info. Perhaps you mean the angle theta is rather small? What do you get for a result? Please show your work.

The answer is at the bottom, and yes the intention is to tell you that $\theta$ is small. Also, I do not know the work to arrive at this answer, that is why I am asking.

 

[PhanthomJay]

There is an approximate but very close relationship between theta and sin or tan theta when theta is small and expressed in radians. But first, you should apply Newton's first law along the direction of the incline after first drawing a free body diagram showing the forces acting along the incline, to see how your answer compares to the book answer.

 

[Nickie]

To find the force of friction, we can use the equation F_f = u*N, where u is the coefficient of static friction and N is the normal force. The normal force is equal to the force of gravity, F_g, multiplied by the cosine of the angle of inclination, \theta. Therefore, the equation for the force of friction becomes F_f = u*F_g*cos(\theta). Since we are looking for the force of friction in terms of \theta and F_g, this is the correct answer. It does not require any trig identities, as the cosine function is already present in the equation.