physics forces question please :(?
Posted on Jun 26, 2009 under snowmobile travel |-Two people are pushing a snowmobile of mass 1500 kg. If they each push with a force of 450 N at an angle of 18 degrees and 300 N at an angle of 15 degrees to the direction the car is travelling, calculate the acceleration of the car (assume no resistance)
- Mark pushes a 6.0 kg lawnmower (the old kind with no engine) with force of 120 N at an angle of 50 degrees to the horizontal. The loss of force due to friction is 60 N.
A. What is the acccleration of the lawnmower?
B. With what force are you pushing down on the mower?
I hope you can help me out. Thank you 
The horizontal component of 450N is 450 cos 18 and that of the 300N is
300 cos 15.
Sum of these two forces is 450 cos 18 + 300 cos 15 =.717.75 N
Acceleration = Force /mass = 717.75 /1500 = 0.4785 m/s^2
===================================
Horizontal force is 120 cos 50 =77.135 N.
Subtracting the loss the net force is 77.135 – 60 = 17.135 N
Acceleration = force / mass = 17.135 / 6= 2.86 m/s^2
===================================
B the force with we push the lawn mover down is
120 sin 50 = 91.93 N
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By edwardare on Jun 26, 2009 | Reply
Use F = ma
If there's an angle involved, then the resultant force is F * cos (the included angle).
If stuck, re-read your text book. If I just spelt out the answer for you, you won't learn it.
References :
By Lisa on Jun 26, 2009 | Reply
Do your own homework please. Just ask your teacher if you are struggling.
References :
By Pearlsawme on Jun 26, 2009 | Reply
The horizontal component of 450N is 450 cos 18 and that of the 300N is
300 cos 15.
Sum of these two forces is 450 cos 18 + 300 cos 15 =.717.75 N
Acceleration = Force /mass = 717.75 /1500 = 0.4785 m/s^2
===================================
Horizontal force is 120 cos 50 =77.135 N.
Subtracting the loss the net force is 77.135 – 60 = 17.135 N
Acceleration = force / mass = 17.135 / 6= 2.86 m/s^2
===================================
B the force with we push the lawn mover down is
120 sin 50 = 91.93 N
References :
By Kelvin Calorie on Jun 26, 2009 | Reply
In your first question, one only cosiders the force applied to the x axis. x axis is the agjacent side of your mentione angles.We know the formula for acceleration in this case is a=f/m. In this case it is the sum of the forces. (cos 18deg x 450N) + (Cos 15 deg x 300n) / 1500Kg.
Your second question should be similar
References :