Where did you get that number?

The first page of the datasheet posted on the page you linked states, "Max 15VDC and 7A" (near the top).

So the maximum input power for the motor is $P=IV$, or $P=(7)(15)=105\mbox{W}$.

Mechanical rotating power can be calculated by $P=\tau \omega $, where $\tau$ is torque in Nm and $\omega $ is rotational speed in rad/s.

The motor's top speed is 10 rpm. Multiply by 2$\pi$ to get to rad/m, then divide by 60 to convert to seconds and you get $\omega=10*(6.28/60) = 1.047 \mbox{rad/s} $.

Now, divide input power by rotational speed to get the theoretical maximum torque: $\tau = 105/1.047 = 100\mbox{Nm} $.

Note that this is Newton meters; aNewton is about a quarter of a pound, so the maximum force is about 25 pounds.

However, an electric motor is generally only about 80% efficient, so I wouldn't expect more than 20 pounds at 1m, max.

Where did you get that number?

The first page of the datasheet posted on the page you linked states, "Max 15VDC and 7A" (near the top).

So the maximum input power for the motor is $P=IV$, or $P=(7)(15)=105\mbox{W}$.

Mechanical rotating power can be calculated by $P=\tau \omega $, where $\tau$ is torque in Nm and $\omega $ is rotational speed in rad/s.

The motor's top speed is 10 rpm. Multiply by 2$\pi$ to get to rad/m, then divide by 60 to convert to seconds and you get $\omega=10*(6.28/60) = 1.047 \mbox{rad/s} $.

Now, divide input power by rotational speed to get the theoretical maximum torque: $\tau = 105/1.047 = 100\mbox{Nm} $.

Note that this is Newton meters; a Newton is about a quarter of a pound, so the maximum force is about 25 pounds.

However, an electric motor is generally only about 80% efficient, so I wouldn't expect more than 20 pounds at 1m, max.