What is the net force acting on the truck?

[lpw8340]

1. A man is lazily floating on an air mattress in a swimming pool. If the weight of the man and air mattress together is 740 N, what is the upward force of the water acting on the system of man and mattress?




2. A truck driving on a level highway is acted upon by the following forces: a downward gravitational force of 48 kN (kiloNewtons); an upward contact force due to the road of 48 kN; another contact force due to the road of 4.5 kN, directed east; and a drag force due to air resistance of 0.9 kN, directed west. What is the net force acting on the truck?
and in what direction is it moving?





3.Vector a has components ax = - 2.9 m/s2 and ay = + 4.1 m/s2.

(a) What is the magnitude of vector a?
____ m/s2

(b) What is the direction of vector a?
____ ° CCW from the +y-axis

 

[blochwave]

Well for 1) you know the guy is floating. In other words, he's not sinking or rising, so the net force on him in the vertical axis is 0. You know the force of gravity on him, it's given. You know the net force is 0. Sooo you know _______ <--take a stab

2)Draw it and use trig

3)Same as above. It's one vector with magnitude 2.9 m/s^2 going left on the x axis, and one going 4.1 m/s^2 up. You can draw a right triangle and use trig

 

[Moetasim]

1. The upward force of the water acting on the system of man and mattress would be equal to 740 N, as the man and mattress are in equilibrium and the net force acting on them is zero. This is due to the fact that the man and mattress are floating on the water, and the water is exerting an upward force on them to counteract the downward force of gravity.

2. The net force acting on the truck would be 3.6 kN directed east. This can be calculated by subtracting the drag force of 0.9 kN from the sum of the gravitational force of 48 kN and the two contact forces of 48 kN and 4.5 kN (48 kN + 48 kN + 4.5 kN - 0.9 kN = 99.6 kN). The truck would be moving in the direction of the net force, which is east.

3. (a) The magnitude of vector a can be calculated using the Pythagorean theorem: √((-2.9)^2 + (4.1)^2) = 5.0 m/s2.

(b) The direction of vector a can be found using trigonometry. The angle θ can be calculated using the inverse tangent function: θ = tan^-1(4.1/-2.9) = -55.0°. This means that vector a is 55.0° counterclockwise from the +y-axis.