Calculating Acceleration of a Box on a Frictionless Surface

[codyM]

Homework Statement

A box of mass 20.00kg placed on a frictionless surface is pulled by two horizontal forces in the plane of the floor, as shown below. Determine the following quantities.
a) acceleration vector.
b) magnitude of resultant acceleration and its angle with the x-axis.

Homework Equations

f=m*a

The Attempt at a Solution

a) (F1-F2)/M=A
f1= 100cos(60)
f2= 200sin(30)
(100cos(60)-200sin(30))/20*9.8=a
a=-25/98

b) I am not sure how to do this.

 

[kuruman]

Can you find the net force acting on the box? Hint: Force is a vector, so you need to add the two forces as vectors.

 

[Doubell]

my idea on the solution for part a. is that the two angles are 60 and 30 outside the horizotal forces this implies that the angle inside should be 90 because the sum of the angles on a straight line =180 as shown by the horizontal dashes on the diagram. by establishing the fact that the angle between the forces is 90 we can use pythagoras theorem to calculate the resultant force acting on the body. the resultant force is the hypothenuse so by squaring the two forces and finding the square root of their sum i got 223 N as the resultant force. F=m *a so dividing the 223 N by the mass of the body gives 11.2 m/s^2 that's my solution.

 

[kuruman]

my idea on the solution for part a. is that the two angles are 60 and 30 outside the horizotal forces this implies that the angle inside should be 90 because the sum of the angles on a straight line =180 as shown by the horizontal dashes on the diagram. by establishing the fact that the angle between the forces is 90 we can use pythagoras theorem to calculate the resultant force acting on the body. the resultant force is the hypothenuse so by squaring the two forces and finding the square root of their sum i got 223 N as the resultant force. F=m *a so dividing the 223 N by the mass of the body gives 11.2 m/s^2 that's my solution.

That is a good idea, Doubell, for a starting point for part (b) because it gives the magnitude of the acceleration vector. How about the angle? What about part (a)? It asks for the acceleration vector. How would one describe the acceleration as a vector?

 

[Doubell]

That is a good idea, Doubell, for a starting point for part (b) because it gives the magnitude of the acceleration vector. How about the angle? What about part (a)? It asks for the acceleration vector. How would one describe the acceleration as a vector?

If by describe u mean define the acceleration as it relates to a vector well a vector posses both magnitude and direction. also vectors can be assigned either positive or negative values and acceleration can be assign + or - values so that's how i describe the acceleration as a vector. for part b the angle with the x-axis would be sin^-1 of 100N/223N which gives (27 degrees[/B]) taking the 100N to be the horzontal force (x axis) and the 200N vertical force (y axis). the hypothenuse already established would be 223N,
magnitude of the acceleration would be 223N at 27 degrees(2 sig fig) to the x axis.

 

[kuruman]

If by describe u mean define the acceleration as it relates to a vector well a vector posses both magnitude and direction. also vectors can be assigned either positive or negative values and acceleration can be assign + or - values so that's how i describe the acceleration as a vector.

Perhaps "describe" was bad wording. How about specify? To specify a vector in 2-d you need to give either its components or a magnitude and an angle.

for part b the angle with the x-axis would be sin^-1 of 100N/223N which gives (27 degrees[/B]) taking the 100N to be the horzontal force (x axis)

You can't do that. The drawing shows the choice of x and y axes and they are not along the force vectors. You have to use the positive x-axis that is given and measure the angle with respect to it.